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CBSE - Class 11 Mathematics Binomial Theorem Worksheet
Expand each of the expressions 2. $(\frac{2}{x} - \frac{x}{2})^5$
Expand each of the expressions 1. $(1-2x)^5$
Worksheet Answers
Solution:
According to the Binomial Theorem, the expansion of $(a + b)^n$ is given by $\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.
For $(x + 1)^6$:
$(x + 1)^6 = \binom{6}{0}x^6 + \binom{6}{1}x^5(1) + \binom{6}{2}x^4(1)^2 + \binom{6}{3}x^3(1)^3 + \binom{6}{4}x^2(1)^4 + \binom{6}{5}x(1)^5 + \binom{6}{6}(1)^6$Calculating the binomial coefficients $\binom{n}{k} = \frac{n!}{k!(n-k)!}$:
$(x + 1)^6 = x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1$ [Equation 1]For $(x - 1)^6$, note that the terms alternate in sign due to $(-1)^k$:
$(x - 1)^6 = \binom{6}{0}x^6 - \binom{6}{1}x^5(1) + \binom{6}{2}x^4(1)^2 - \binom{6}{3}x^3(1)^3 + \binom{6}{4}x^2(1)^4 - \binom{6}{5}x(1)^5 + \binom{6}{6}(1)^6$ $(x - 1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1$ [Equation 2]Adding [Equation 1] and [Equation 2]:
$(x + 1)^6 + (x - 1)^6 = (x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1) + (x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1)$Observe that odd-powered terms ($x^5, x^3, x$) cancel out [By Algebraic Additive Inverse Property]:
$(x + 1)^6 + (x - 1)^6 = 2(x^6 + 15x^4 + 15x^2 + 1)$Substitute $x = \sqrt{2}$ into the derived formula $2(x^6 + 15x^4 + 15x^2 + 1)$:
Substitute these values:
Value $= 2(8 + 15(4) + 15(2) + 1)$ Value $= 2(8 + 60 + 30 + 1)$ Value $= 2(99)$ Value $= 198$Final Solution: (x + 1)^6 + (x - 1)^6 = 2(x^6 + 15x^4 + 15x^2 + 1); Value = 198
Solution:
We are required to expand the binomial expression $\left(\frac{2}{x} - \frac{x}{2}\right)^5$ using the Binomial Theorem. The standard form for the expansion of $(a + b)^n$ is given by:
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ [Per the Binomial Theorem]
In this specific problem:
Substituting the values into the expansion formula:
$\left(\frac{2}{x} - \frac{x}{2}\right)^5 = \binom{5}{0} \left(\frac{2}{x}\right)^5 + \binom{5}{1} \left(\frac{2}{x}\right)^4 \left(-\frac{x}{2}\right)^1 + \binom{5}{2} \left(\frac{2}{x}\right)^3 \left(-\frac{x}{2}\right)^2 + \binom{5}{3} \left(\frac{2}{x}\right)^2 \left(-\frac{x}{2}\right)^3 + \binom{5}{4} \left(\frac{2}{x}\right)^1 \left(-\frac{x}{2}\right)^4 + \binom{5}{5} \left(-\frac{x}{2}\right)^5$First, calculate the combinations $\binom{n}{k} = \frac{n!}{k!(n-k)!}$:
Expand each term individually:
| Term | Expression | Simplified |
|---|---|---|
| 1st | $1 \cdot \frac{32}{x^5}$ | $\frac{32}{x^5}$ |
| 2nd | $5 \cdot \frac{16}{x^4} \cdot (-\frac{x}{2})$ | $-\frac{40}{x^3}$ |
| 3rd | $10 \cdot \frac{8}{x^3} \cdot \frac{x^2}{4}$ | $\frac{20}{x}$ |
| 4th | $10 \cdot \frac{4}{x^2} \cdot (-\frac{x^3}{8})$ | $-5x$ |
| 5th | $5 \cdot \frac{2}{x} \cdot \frac{x^4}{16}$ | $\frac{5x^3}{8}$ |
| 6th | $1 \cdot (-\frac{x^5}{32})$ | $-\frac{x^5}{32}$ |
Combining all the simplified terms, we obtain:
$\left(\frac{2}{x} - \frac{x}{2}\right)^5 = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}$
Final Solution: \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}
Solution:
To find the value of $(a + b)^4 - (a - b)^4$, we employ the Binomial Theorem, which states that $(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$.
Expanding $(a + b)^4$:
$(a + b)^4 = \binom{4}{0}a^4b^0 + \binom{4}{1}a^3b^1 + \binom{4}{2}a^2b^2 + \binom{4}{3}a^1b^3 + \binom{4}{4}a^0b^4$ $(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$ [By Binomial Coefficient Calculation]Expanding $(a - b)^4$:
$(a - b)^4 = \binom{4}{0}a^4(-b)^0 + \binom{4}{1}a^3(-b)^1 + \binom{4}{2}a^2(-b)^2 + \binom{4}{3}a^1(-b)^3 + \binom{4}{4}a^0(-b)^4$ $(a - b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$ [Per the property of alternating signs in alternating binomial expansions]Subtracting the second expression from the first:
$(a + b)^4 - (a - b)^4 = (a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4) - (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4)$Distributing the negative sign and grouping like terms:
$= a^4 - a^4 + 4a^3b + 4a^3b + 6a^2b^2 - 6a^2b^2 + 4ab^3 + 4ab^3 + b^4 - b^4$ $= 8a^3b + 8ab^3$ $= 8ab(a^2 + b^2)$Given the expression $(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4$, we identify the variables as:
$a = \sqrt{3}$ $b = \sqrt{2}$Substituting into the derived identity $8ab(a^2 + b^2)$:
$= 8(\sqrt{3})(\sqrt{2})((\sqrt{3})^2 + (\sqrt{2})^2)$Computing the individual terms:
Final calculation:
$= 8(\sqrt{6})(3 + 2)$ $= 8(\sqrt{6})(5)$ $= 40\sqrt{6}$Final Solution: (a + b)^4 – (a – b)^4 = 8ab(a^2 + b^2); therefore, (\sqrt{3} + \sqrt{2})^4 – (\sqrt{3} – \sqrt{2})^4 = 40\sqrt{6}
Solution:
To expand the expression $(1 + \frac{x}{2} - \frac{2}{x})^{4}$, we treat the trinomial as a binomial by grouping terms: $[1 + (\frac{x}{2} - \frac{2}{x})]^{4}$.
Per the Binomial Theorem, for any positive integer $n$:
$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$Setting $a = 1$, $b = (\frac{x}{2} - \frac{2}{x})$, and $n = 4$:
$(1 + \frac{x}{2} - \frac{2}{x})^4 = \binom{4}{0}(1)^4 + \binom{4}{1}(1)^3(\frac{x}{2} - \frac{2}{x}) + \binom{4}{2}(1)^2(\frac{x}{2} - \frac{2}{x})^2 + \binom{4}{3}(1)^1(\frac{x}{2} - \frac{2}{x})^3 + \binom{4}{4}(\frac{x}{2} - \frac{2}{x})^4$We calculate the binomial coefficients: $\binom{4}{0}=1, \binom{4}{1}=4, \binom{4}{2}=6, \binom{4}{3}=4, \binom{4}{4}=1$.
Summing all components calculated in Step 2:
| Power of x | Summation |
|---|---|
| $x^4$ | $\frac{x^4}{16}$ |
| $x^3$ | $\frac{x^3}{2}$ |
| $x^2$ | $\frac{3x^2}{2} - x^2 = \frac{x^2}{2}$ |
| $x^1$ | $2x - 6x = -4x$ |
| Constant | $1 - 12 + 6 = -5$ |
| $x^{-1}$ | $-\frac{8}{x} + \frac{24}{x} = \frac{16}{x}$ |
| $x^{-2}$ | $\frac{24}{x^2} - \frac{16}{x^2} = \frac{8}{x^2}$ |
| $x^{-3}$ | $-\frac{32}{x^3}$ |
| $x^{-4}$ | $\frac{16}{x^4}$ |
Final Solution: \frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x - 5 + \frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4}
Solution:
To prove the identity $\sum_{r=0}^{n} {^nC_r} 3^r = 4^n$, we utilize the fundamental Binomial Theorem for any positive integer $n$.
The general form of the Binomial Theorem is expressed as:
$(x + a)^n = \sum_{r=0}^{n} {^nC_r} x^{n-r} a^r$ [Per the Binomial Theorem]
In the given expression $\sum_{r=0}^{n} {^nC_r} 3^r$, we compare the terms with the standard expansion of $(x + a)^n$.
Substituting these values into the standard expansion formula:
$(1 + 3)^n = \sum_{r=0}^{n} {^nC_r} (1)^{n-r} (3)^r$
Since $(1)^{n-r} = 1$ for any non-negative integer $r$ such that $r \leq n$, the expression simplifies significantly [By the definition of exponentiation]:
$(1 + 3)^n = \sum_{r=0}^{n} {^nC_r} \cdot 1 \cdot 3^r$
$(1 + 3)^n = \sum_{r=0}^{n} {^nC_r} 3^r$
Perform the arithmetic inside the parentheses on the left-hand side of the equation:
$4^n = \sum_{r=0}^{n} {^nC_r} 3^r$
Or, expressed in the requested order:
$\sum_{r=0}^{n} {^nC_r} 3^r = 4^n$
Final Solution: The equality $\sum_{r=0}^{n} {^nC_r} 3^r = 4^n$ is verified as a direct application of the Binomial Theorem $(1+x)^n = \sum {^nC_r} x^r$ where $x=3$.
Solution:
To prove that an expression $E(n) = 9^{n+1} - 8n - 9$ is divisible by $64$ for all positive integers $n$, we utilize the Binomial Theorem, which states:
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n}b^n$
[By Binomial Theorem expansion]
We rewrite the base $9$ as $(1 + 8)$. Thus, the term $9^{n+1}$ can be expressed as:
$9^{n+1} = (1 + 8)^{n+1}$
Applying the theorem where $a=1$, $b=8$, and the exponent is $n+1$:
$(1 + 8)^{n+1} = \binom{n+1}{0}(1)^{n+1} + \binom{n+1}{1}(1)^{n}(8) + \binom{n+1}{2}(1)^{n-1}(8)^2 + \binom{n+1}{3}(8)^3 + \dots + \binom{n+1}{n+1}(8)^{n+1}$
[By Binomial Expansion Formula: $(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$]
Evaluating the first two terms of the expansion:
Substituting these values:
$9^{n+1} = 1 + (n+1)(8) + \binom{n+1}{2}(64) + \binom{n+1}{3}(8^3) + \dots + 8^{n+1}$
$9^{n+1} = 1 + 8n + 8 + 64 \left[ \binom{n+1}{2} + \binom{n+1}{3}(8) + \dots + 8^{n-1} \right]$
$9^{n+1} = 9 + 8n + 64 \times K$
where $K = \sum_{k=2}^{n+1} \binom{n+1}{k} 8^{k-2}$ is an integer constant.
Now, substitute this back into the original expression $E(n)$:
$E(n) = 9^{n+1} - 8n - 9$
$E(n) = (9 + 8n + 64K) - 8n - 9$
[Associative and Commutative properties of addition]
$E(n) = 9 - 9 + 8n - 8n + 64K$
$E(n) = 64K$
Since $E(n)$ is equal to $64$ times an integer $K$, $E(n)$ is by definition divisible by $64$.
Final Solution: Since $9^{n+1} - 8n - 9 = 64K$, the expression is divisible by 64 for all positive integers $n$.
Solution:
To expand the binomial expression $(a + b)^n$, we employ the Binomial Theorem, which states: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ [Per the Binomial Theorem for a positive integer index $n$].
Given the expression $(1 - 2x)^5$, we identify the components:
Using the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, we calculate the coefficients for $n=5$:
Substituting the variables into the general expansion formula: $(1 - 2x)^5 = \binom{5}{0}(1)^5(-2x)^0 + \binom{5}{1}(1)^4(-2x)^1 + \binom{5}{2}(1)^3(-2x)^2 + \binom{5}{3}(1)^2(-2x)^3 + \binom{5}{4}(1)^1(-2x)^4 + \binom{5}{5}(1)^0(-2x)^5$
Final Solution: (1 - 2x)^5 = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5
Solution:
To evaluate $(102)^5$ using the Binomial Theorem, we express the base $102$ as the sum of two terms: $100 + 2$. This allows the application of the binomial expansion formula: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ [Per the Binomial Theorem for positive integers].
Here, $a = 100$, $b = 2$, and $n = 5$.
Substituting the values into the binomial formula:
$(100 + 2)^5 = \binom{5}{0}(100)^5(2)^0 + \binom{5}{1}(100)^4(2)^1 + \binom{5}{2}(100)^3(2)^2 + \binom{5}{3}(100)^2(2)^3 + \binom{5}{4}(100)^1(2)^4 + \binom{5}{5}(100)^0(2)^5$
We evaluate each term individually using the definition of combinations $\binom{n}{k} = \frac{n!}{k!(n-k)!}$:
| Term | Coefficient | Power Calculation | Result |
|---|---|---|---|
| $k=0$ | $\binom{5}{0} = 1$ | $1 \cdot 100^5 \cdot 1$ | $10,000,000,000$ |
| $k=1$ | $\binom{5}{1} = 5$ | $5 \cdot 100^4 \cdot 2$ | $1,000,000,000$ |
| $k=2$ | $\binom{5}{2} = 10$ | $10 \cdot 100^3 \cdot 4$ | $40,000,000$ |
| $k=3$ | $\binom{5}{3} = 10$ | $10 \cdot 100^2 \cdot 8$ | $800,000$ |
| $k=4$ | $\binom{5}{4} = 5$ | $5 \cdot 100^1 \cdot 16$ | $8,000$ |
| $k=5$ | $\binom{5}{5} = 1$ | $1 \cdot 1 \cdot 32$ | $32$ |
Sum the calculated values sequentially:
Final Solution: (102)^5 = 11,040,808,032
Solution:
To expand an expression of the form $(a + b)^n$, we employ the Binomial Theorem, which states:
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$
[Per the Binomial Theorem for a positive integer index $n$].
In the given expression $(1 - 2x)^5$, we identify the variables as follows:
Applying the values to the summation formula:
$(1 - 2x)^5 = \binom{5}{0}(1)^5(-2x)^0 + \binom{5}{1}(1)^4(-2x)^1 + \binom{5}{2}(1)^3(-2x)^2 + \binom{5}{3}(1)^2(-2x)^3 + \binom{5}{4}(1)^1(-2x)^4 + \binom{5}{5}(1)^0(-2x)^5$
Using the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$:
| Term ($k$) | Coefficient $\binom{5}{k}$ |
|---|---|
| $k=0$ | $\frac{5!}{0!5!} = 1$ |
| $k=1$ | $\frac{5!}{1!4!} = 5$ |
| $k=2$ | $\frac{5!}{2!3!} = 10$ |
| $k=3$ | $\frac{5!}{3!2!} = 10$ |
| $k=4$ | $\frac{5!}{4!1!} = 5$ |
| $k=5$ | $\frac{5!}{5!0!} = 1$ |
Substituting the coefficients and evaluating the powers:
Combining all terms, we obtain the full expanded polynomial:
$(1 - 2x)^5 = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$
Final Solution: $1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$
Solution:
To evaluate $(102)^5$ using the Binomial Theorem, we express the base as a sum of two terms: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.
Here, $a = 100$, $b = 2$, and $n = 5$. Thus:
$(100 + 2)^5 = \sum_{k=0}^{5} \binom{5}{k} (100)^{5-k} (2)^k$[Per the Binomial Theorem for positive integer indices]
Expanding the summation into its individual components:
$(102)^5 = \binom{5}{0}(100)^5(2)^0 + \binom{5}{1}(100)^4(2)^1 + \binom{5}{2}(100)^3(2)^2 + \binom{5}{3}(100)^2(2)^3 + \binom{5}{4}(100)^1(2)^4 + \binom{5}{5}(100)^0(2)^5$
Using the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, we determine the coefficients:
| Term | Calculation | Value |
|---|---|---|
| $k=0$ | $1 \times 10,000,000,000 \times 1$ | $10,000,000,000$ |
| $k=1$ | $5 \times 100,000,000 \times 2$ | $1,000,000,000$ |
| $k=2$ | $10 \times 1,000,000 \times 4$ | $40,000,000$ |
| $k=3$ | $10 \times 10,000 \times 8$ | $800,000$ |
| $k=4$ | $5 \times 100 \times 16$ | $8,000$ |
| $k=5$ | $1 \times 1 \times 32$ | $32$ |
Summing the resulting values:
$10,000,000,000 + 1,000,000,000 + 40,000,000 + 800,000 + 8,000 + 32$Final Solution: 11,040,808,032
Solution:
To expand the expression $\left(\frac{2}{x} - \frac{x}{2}\right)^5$, we utilize the Binomial Theorem, which states that for any positive integer $n$:
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$
In this expression:
Substituting the values into the binomial formula:
$\left(\frac{2}{x} - \frac{x}{2}\right)^5 = \binom{5}{0} \left(\frac{2}{x}\right)^5 + \binom{5}{1} \left(\frac{2}{x}\right)^4 \left(-\frac{x}{2}\right)^1 + \binom{5}{2} \left(\frac{2}{x}\right)^3 \left(-\frac{x}{2}\right)^2 + \binom{5}{3} \left(\frac{2}{x}\right)^2 \left(-\frac{x}{2}\right)^3 + \binom{5}{4} \left(\frac{2}{x}\right)^1 \left(-\frac{x}{2}\right)^4 + \binom{5}{5} \left(-\frac{x}{2}\right)^5$
We evaluate the binomial coefficients $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ [By definition of combinations]:
| Coefficient | Value |
|---|---|
| $\binom{5}{0}$ | $1$ |
| $\binom{5}{1}$ | $5$ |
| $\binom{5}{2}$ | $10$ |
| $\binom{5}{3}$ | $10$ |
| $\binom{5}{4}$ | $5$ |
| $\binom{5}{5}$ | $1$ |
Substitute the coefficients and expand the powers:
Combining all the simplified terms into a single expression:
$\left(\frac{2}{x} - \frac{x}{2}\right)^5 = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}$
Final Solution: \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}
Solution:
We are required to compare the values of $x = (1.1)^{10000}$ and $y = 1000$ using the Binomial Theorem.
The Binomial Theorem states that for any positive integer $n$:
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$
Setting $a = 1$, $b = 0.1$, and $n = 10000$, we expand $(1 + 0.1)^{10000}$:
$(1 + 0.1)^{10000} = \binom{10000}{0}(1)^{10000}(0.1)^0 + \binom{10000}{1}(1)^{9999}(0.1)^1 + \binom{10000}{2}(1)^{9998}(0.1)^2 + \dots$
[By the Binomial Theorem expansion protocol]
We calculate the first few terms of the series:
The total value is the sum of all terms from $k=0$ to $k=10000$:
$(1.1)^{10000} = 1 + 1000 + 499950 + (\text{all subsequent positive terms})$
Summing the first three terms alone:
$1 + 1000 + 499950 = 500951$
Since $(1.1)^{10000} = 500951 + \text{positive terms}$, it is evident that:
$(1.1)^{10000} > 500951$
Given the target value $1000$, we observe:
$500951 > 1000$
[By the Transitive Property of Inequality, if $A > B$ and $B > C$, then $A > C$]
Final Solution: $(1.1)^{10000}$ is significantly larger than $1000$.
Solution:
To evaluate $(101)^4$ using the Binomial Theorem, we express the base $101$ as the sum of two terms, $100$ and $1$, such that the calculation is simplified through powers of $10$.
Expression: $(101)^4 = (100 + 1)^4$
The Binomial Theorem states that for any positive integer $n$:
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ [Per the Binomial Theorem]Here, $a = 100$, $b = 1$, and $n = 4$. Substituting these values:
$(100 + 1)^4 = \binom{4}{0}(100)^4(1)^0 + \binom{4}{1}(100)^3(1)^1 + \binom{4}{2}(100)^2(1)^2 + \binom{4}{3}(100)^1(1)^3 + \binom{4}{4}(100)^0(1)^4$We calculate the combinations using the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$:
Substituting the coefficients and powers into the expansion:
$(100 + 1)^4 = 1(100,000,000) + 4(1,000,000) + 6(10,000) + 4(100) + 1(1)$| 100,000,000 |
| 4,000,000 |
| 60,000 |
| 400 |
| 1 |
| 104,060,401 |
Final Solution: 104,060,401
Solution:
To find the expansion of $(3x^2 - 2ax + 3a^2)^3$, we utilize the Binomial Theorem, which states $(A + B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k} B^k$. Since the expression is a trinomial, we group it into a binomial form: $[(3x^2 + 3a^2) - 2ax]^3$.
Let $A = (3x^2 + 3a^2)$ and $B = -2ax$. Per the Binomial Theorem for $n=3$:
$(A + B)^3 = 1 \cdot A^3 + 3 \cdot A^2 \cdot B + 3 \cdot A \cdot B^2 + 1 \cdot B^3$
Substituting $A = 3(x^2 + a^2)$ and $B = -2ax$:
Using $(x^2+a^2)^3 = (x^2)^3 + 3(x^2)^2(a^2) + 3(x^2)(a^2)^2 + (a^2)^3 = x^6 + 3x^4a^2 + 3x^2a^4 + a^6$.
Result: $27x^6 + 81x^4a^2 + 81x^2a^4 + 27a^6$.
$= 27(x^4 + 2x^2a^2 + a^4) \cdot (-2ax) = -54ax^5 - 108a^3x^3 - 54a^5x$.
$= 36a^2x^2(x^2+a^2) = 36x^4a^2 + 36x^2a^4$.
Summing the calculated terms:
| Component | Algebraic Expression |
|---|---|
| $A^3$ | $27x^6 + 81x^4a^2 + 81x^2a^4 + 27a^6$ |
| $3A^2B$ | $-54x^5a - 108x^3a^3 - 54xa^5$ |
| $3AB^2$ | $36x^4a^2 + 36x^2a^4$ |
| $B^3$ | $-8x^3a^3$ |
Combining like terms by degree:
Final Solution: $27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6$
Solution:
To evaluate $(99)^5$, we express the base as a binomial sum or difference to facilitate calculation using the Binomial Theorem. The expression $(99)^5$ is equivalent to $(100 - 1)^5$.
The Binomial Theorem states that for any positive integer $n$:
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$
Here, $a = 100$, $b = -1$, and $n = 5$. [By the Binomial Theorem expansion rule]
Expanding $(100 - 1)^5$ yields the following terms:
$(100 - 1)^5 = \binom{5}{0}(100)^5(-1)^0 + \binom{5}{1}(100)^4(-1)^1 + \binom{5}{2}(100)^3(-1)^2 + \binom{5}{3}(100)^2(-1)^3 + \binom{5}{4}(100)^1(-1)^4 + \binom{5}{5}(100)^0(-1)^5$Calculating each coefficient and power individually:
The final value is determined by the algebraic sum of the terms calculated in Step 3:
Sum $= 10,000,000,000 - 500,000,000 + 10,000,000 - 100,000 + 500 - 1$
| Positive Terms | Negative Terms |
|---|---|
| $10,000,000,000$ | $500,000,000$ |
| $10,000,000$ | $100,000$ |
| $500$ | $1$ |
| Total: $10,010,000,500$ | Total: $500,100,001$ |
Performing the final subtraction: $10,010,000,500 - 500,100,001 = 9,509,900,499$
Final Solution: (99)^5 = 9,509,900,499
Solution:
To prove that $9^{n+1} - 8n - 9$ is divisible by $64$, we must express $9^{n+1}$ in a form conducive to the Binomial Theorem. We rewrite $9$ as $(1 + 8)$.
According to the Binomial Theorem, for any positive integer $m$:
$(a + b)^m = \sum_{k=0}^{m} \binom{m}{k} a^{m-k} b^k$Setting $a=1$, $b=8$, and $m=n+1$, we obtain:
$9^{n+1} = (1 + 8)^{n+1} = \binom{n+1}{0}(1)^{n+1} + \binom{n+1}{1}(1)^n(8)^1 + \binom{n+1}{2}(1)^{n-1}(8)^2 + \binom{n+1}{3}(8)^3 + \dots + \binom{n+1}{n+1}(8)^{n+1}$[By Binomial Theorem expansion]
We evaluate the first few terms of the expansion:
Substituting these back into the expansion:
$9^{n+1} = 1 + (8n + 8) + \sum_{k=2}^{n+1} \binom{n+1}{k} 8^k$ $9^{n+1} = 9 + 8n + \sum_{k=2}^{n+1} \binom{n+1}{k} 8^k$We now isolate the expression given in the problem statement by subtracting $(8n + 9)$ from both sides:
$9^{n+1} - 8n - 9 = \sum_{k=2}^{n+1} \binom{n+1}{k} 8^k$Focusing on the right-hand side summation, we extract the factor $8^2 = 64$:
$\sum_{k=2}^{n+1} \binom{n+1}{k} 8^k = 64 \left[ \binom{n+1}{2} + \binom{n+1}{3} \cdot 8 + \dots + \binom{n+1}{n+1} 8^{n-1} \right]$[By distributive property of multiplication]
Since the summation $\left[ \binom{n+1}{2} + \binom{n+1}{3} \cdot 8 + \dots + \binom{n+1}{n+1} 8^{n-1} \right]$ results in an integer value (as binomial coefficients and powers of integers are integers), let this sum be represented by the integer $I$.
Thus, $9^{n+1} - 8n - 9 = 64I$.
Because the expression is equal to $64$ multiplied by an integer $I$, it follows by the definition of divisibility that the expression is divisible by $64$.
Final Solution: Since the expansion results in $64 \times I$ where $I \in \mathbb{Z}$, the expression $(9^{n+1} – 8n – 9)$ is proven to be divisible by 64 for all positive integers $n$.
Solution:
To evaluate $(99)^5$ using the Binomial Theorem, we first express the base $99$ as a difference of two numbers such that the binomial expansion is simplified. We define $99 = (100 - 1)$.
The Binomial Theorem states: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$
Setting $a = 100$, $b = -1$, and $n = 5$, we get:
$(100 - 1)^5 = \sum_{k=0}^{5} \binom{5}{k} (100)^{5-k} (-1)^k$ [By Binomial Theorem Expansion]
We expand the summation for $k = 0$ to $5$:
| Term ($k$) | Calculation | Result |
|---|---|---|
| $k=0$ | $\binom{5}{0} (100)^5 (-1)^0$ | $1 \times 10,000,000,000 \times 1 = 10,000,000,000$ |
| $k=1$ | $\binom{5}{1} (100)^4 (-1)^1$ | $5 \times 100,000,000 \times (-1) = -500,000,000$ |
| $k=2$ | $\binom{5}{2} (100)^3 (-1)^2$ | $10 \times 1,000,000 \times 1 = 10,000,000$ |
| $k=3$ | $\binom{5}{3} (100)^2 (-1)^3$ | $10 \times 10,000 \times (-1) = -100,000$ |
| $k=4$ | $\binom{5}{4} (100)^1 (-1)^4$ | $5 \times 100 \times 1 = 500$ |
| $k=5$ | $\binom{5}{5} (100)^0 (-1)^5$ | $1 \times 1 \times (-1) = -1$ |
Summing the calculated terms:
$S = 10,000,000,000 - 500,000,000 + 10,000,000 - 100,000 + 500 - 1$
Grouping positive and negative terms:
Positive terms: $10,000,000,000 + 10,000,000 + 500 = 10,010,000,500$
Negative terms: $-500,000,000 - 100,000 - 1 = -500,100,001$
$S = 10,010,000,500 - 500,100,001$
$10,010,000,500 - 500,100,001 = 9,509,900,499$
Final Solution: 9,509,900,499
Solution:
To evaluate $(101)^4$ using the Binomial Theorem, we express the base $101$ as a sum of two terms, $100$ and $1$, such that the calculation is simplified through powers of $10$.
Let $x = 100$ and $a = 1$. Then the expression becomes:
$(101)^4 = (100 + 1)^4$
Per the Binomial Theorem, the expansion of $(x + a)^n$ is defined as:
$(x + a)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k$
Where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ [By definition of Binomial Coefficients].
Setting $n = 4$, $x = 100$, and $a = 1$, the expansion is:
$(100 + 1)^4 = \binom{4}{0}(100)^4(1)^0 + \binom{4}{1}(100)^3(1)^1 + \binom{4}{2}(100)^2(1)^2 + \binom{4}{3}(100)^1(1)^3 + \binom{4}{4}(100)^0(1)^4$
We calculate the coefficients $\binom{n}{k}$ for $n=4$:
Substituting the coefficients and powers into the expansion:
$(101)^4 = 1(100,000,000) + 4(1,000,000) + 6(10,000) + 4(100) + 1(1)$
Performing the multiplication:
$(101)^4 = 100,000,000 + 4,000,000 + 60,000 + 400 + 1
Adding the components:
| $100,000,000$ |
| $4,000,000$ |
| $60,000$ |
| $400$ |
| $1$ |
| $104,060,401$ |
Final Solution: $(101)^4 = 104,060,401$
Solution:
To evaluate the summation $\sum_{r=0}^{n} {}^{n}C_{r} 3^{r}$, we reference the Binomial Theorem, which states that for any positive integer $n$ and any real numbers $a$ and $b$:
$(a + b)^{n} = \sum_{r=0}^{n} {}^{n}C_{r} a^{n-r} b^{r}$
[Per the Binomial Theorem definition for non-negative integral exponents].
To match the structure of the given expression $\sum_{r=0}^{n} {}^{n}C_{r} 3^{r}$, we perform the following variable assignments within the general expansion:
Given that $(1)^{n-r} = 1$ for all non-negative integers $n$ and $r$ where $n \geq r$ [By the property of unity in exponentiation]:
$(1 + 3)^{n} = \sum_{r=0}^{n} {}^{n}C_{r} (1) (3)^{r}$
$(4)^{n} = \sum_{r=0}^{n} {}^{n}C_{r} 3^{r}$
The LHS (Left Hand Side) of the expansion simplifies directly to the sum of the coefficients multiplied by the powers of the variable $3$, resulting in the power of the binomial sum $(1+3)$. Thus, we have verified that:
$\sum_{r=0}^{n} {}^{n}C_{r} 3^{r} = 4^{n}$
Final Solution: The identity $\sum_{r=0}^{n} {}^{n}C_{r} 3^{r} = 4^{n}$ is proven.
Solution:
To expand the expression $\left(\frac{x}{3} + \frac{1}{x}\right)^{5}$, we utilize the Binomial Theorem, which states that for any non-negative integer $n$:
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$Where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
In this problem, we identify the following variables: $a = \frac{x}{3}$, $b = \frac{1}{x}$, and $n = 5$.
Using Pascal’s triangle or the combination formula for $n=5$, the coefficients are:
Substituting the values into the binomial expansion formula:
$\left(\frac{x}{3} + \frac{1}{x}\right)^{5} = \binom{5}{0} \left(\frac{x}{3}\right)^5 + \binom{5}{1} \left(\frac{x}{3}\right)^4 \left(\frac{1}{x}\right)^1 + \binom{5}{2} \left(\frac{x}{3}\right)^3 \left(\frac{1}{x}\right)^2 + \binom{5}{3} \left(\frac{x}{3}\right)^2 \left(\frac{1}{x}\right)^3 + \binom{5}{4} \left(\frac{x}{3}\right)^1 \left(\frac{1}{x}\right)^4 + \binom{5}{5} \left(\frac{1}{x}\right)^5$Executing the power operations for each term [By Law of Exponents $(ab)^n = a^n b^n$]:
| Term | Expansion | Result |
|---|---|---|
| 1st | $1 \cdot \frac{x^5}{243}$ | $\frac{x^5}{243}$ |
| 2nd | $5 \cdot \frac{x^4}{81} \cdot \frac{1}{x}$ | $\frac{5x^3}{81}$ |
| 3rd | $10 \cdot \frac{x^3}{27} \cdot \frac{1}{x^2}$ | $\frac{10x}{27}$ |
| 4th | $10 \cdot \frac{x^2}{9} \cdot \frac{1}{x^3}$ | $\frac{10}{9x}$ |
| 5th | $5 \cdot \frac{x}{3} \cdot \frac{1}{x^4}$ | $\frac{5}{3x^3}$ |
| 6th | $1 \cdot \frac{1}{x^5}$ | $\frac{1}{x^5}$ |
Combining all simplified terms yields the final expansion:
Final Solution: $\frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$
