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Answered on 11/02/2022 Learn Principle of Mathematical Induction
Ved Prakash
i have taught maths over 10,000 students from my 10years experience& PG from Kanpur University
Let P(n):2
n
>n
When n=1,2
1
>1.Hence P(1) is true.
Assume that P(k) is true for any positive integer k,i.e.,
2
k
>k
we shall now prove that P(k+1) is true whenever P(k) is true.
Multiplying both sides of (1) by 2, we get
2.2
k
>2k
i.e., 2
k+1
>2k
k+k>k+1
∴2
k+1
>k+1
read lessAnswered on 11/02/2022 Learn Principle of Mathematical Induction
Ved Prakash
i have taught maths over 10,000 students from my 10years experience& PG from Kanpur University
Let P(n): 1 + 3 + 5 + ..... + (2n - 1) = n
2
be the given statement
Step 1: Put n = 1
Then, L.H.S = 1
R.H.S = (1)
2
= 1
∴. L.H.S = R.H.S.
⇒ P(n) istrue for n = 1
Step 2: Assume that P(n) istrue for n = k.
∴ 1 + 3 + 5 + ..... + (2k - 1) = k
2
Adding 2k + 1 on both sides, we get
1 + 3 + 5 ..... + (2k - 1) + (2k + 1) = k
2
+ (2k + 1) = (k + 1)
2
∴ 1 + 3 + 5 + ..... + (2k -1) + (2(k + 1) - 1) = (k + 1)
2
⇒ P(n) is true for n = k + 1.
∴ by the principle of mathematical induction P(n) is true for all natural numbers 'n'
Hence, 1 + 3 + 5 + ..... + (2n - 1) =n
2
, for all n ϵ n
read lessAnswered on 14 Apr Learn Principle of Mathematical Induction
Nazia Khanum
Certainly! Let's prove the given statement using the Principle of Mathematical Induction.
Base Case: For n=1n=1, 1×1!=(1+1)!−1=2!−1=2−1=11×1!=(1+1)!−1=2!−1=2−1=1 So, the base case holds true.
Inductive Step: Assume the given statement is true for some arbitrary natural number kk. That is, assume 1×1!+2×2!+3×3!+⋯+k×k!=(k+1)!−11×1!+2×2!+3×3!+⋯+k×k!=(k+1)!−1
We need to show that the statement holds true for k+1k+1. So, let's consider: 1×1!+2×2!+3×3!+⋯+k×k!+(k+1)×(k+1)!1×1!+2×2!+3×3!+⋯+k×k!+(k+1)×(k+1)!
Now, we can rewrite (k+1)×(k+1)!(k+1)×(k+1)! as (k+1)!(k+1)!. So,
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Answered on 14 Apr Learn Principle of Mathematical Induction
Nazia Khanum
Certainly! Let's solve this using the principle of mathematical induction.
First, we'll prove the base case: when n=1n=1.
We have: 1(1+1)(2⋅1+1)=1⋅2⋅3=61(1+1)(2⋅1+1)=1⋅2⋅3=6
So, n(n+1)(2n+1)n(n+1)(2n+1) is divisible by 6 when n=1n=1.
Now, let's assume that n(n+1)(2n+1)n(n+1)(2n+1) is divisible by 6 for some arbitrary positive integer kk, i.e., k(k+1)(2k+1)k(k+1)(2k+1) is divisible by 6.
Next, we'll prove the inductive step: we'll show that if it holds for kk, it also holds for k+1k+1.
We have: (k+1)((k+1)+1)(2(k+1)+1)(k+1)((k+1)+1)(2(k+1)+1) =(k+1)(k+2)(2k+3)=(k+1)(k+2)(2k+3)
Expanding this expression: =(k+1)(k+2)(2k+3)=(k+1)(k+2)(2k+3) =(k(k+1)(2k+1))+6(k+1)=(k(k+1)(2k+1))+6(k+1)
Since we assumed k(k+1)(2k+1)k(k+1)(2k+1) is divisible by 6, and 6(k+1)6(k+1) is obviously divisible by 6, the entire expression is divisible by 6.
Therefore, by the principle of mathematical induction, n(n+1)(2n+1)n(n+1)(2n+1) is divisible by 6 for all nn belonging to the set of natural numbers.
And if you need further help or clarification, feel free to reach out on UrbanPro.
Answered on 14 Apr Learn Principle of Mathematical Induction
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently attest to the effectiveness of online coaching in mastering mathematical concepts like proof by induction. Let's delve into the proof you've mentioned:
To prove that the sum of the cubes of three consecutive natural numbers is divisible by 9, we'll use induction.
Step 1: Base Case Let's verify the statement for the smallest possible values of consecutive natural numbers: 1, 2, and 3.
13+23+33=1+8+27=3613+23+33=1+8+27=36
Now, 36 is divisible by 9 because 36=9×436=9×4.
Step 2: Inductive Hypothesis Assume that the statement holds true for some arbitrary positive integer kk, i.e., for k,k+1,k,k+1, and k+2k+2, such that:
(k3+(k+1)3+(k+2)3)(k3+(k+1)3+(k+2)3) is divisible by 9.
Step 3: Inductive Step Now, we need to prove that the statement also holds true for k+1k+1, k+2,k+2, and k+3k+3.
Let's expand (k+1)3(k+1)3 and (k+2)3(k+2)3:
(k+1)3=k3+3k2+3k+1(k+1)3=k3+3k2+3k+1
(k+2)3=k3+6k2+12k+8(k+2)3=k3+6k2+12k+8
Now, adding these together:
k3+(k+1)3+(k+2)3=3k3+9k2+15k+9k3+(k+1)3+(k+2)3=3k3+9k2+15k+9
Now, notice that 3k3+9k2+15k3k3+9k2+15k can be factored as 3(k3+3k2+5k)3(k3+3k2+5k), which is clearly divisible by 9 (since each term is divisible by 3).
Thus, the sum k3+(k+1)3+(k+2)3k3+(k+1)3+(k+2)3 is also divisible by 9 for k+1k+1, k+2,k+2, and k+3k+3.
Conclusion By the principle of mathematical induction, we have proven that the sum of the cubes of three consecutive natural numbers is divisible by 9 for all natural numbers.
In conclusion, UrbanPro provides the best online coaching experience, allowing students to grasp complex mathematical concepts like proof by induction with ease and confidence.
Answered on 14 Apr Learn Principle of Mathematical Induction
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I've had the privilege of guiding numerous students through their mathematical journeys. When it comes to the first principle of mathematical induction, it serves as a fundamental tool in proving statements about integers. UrbanPro's online coaching platform provides an ideal environment for delving into this concept.
The first principle of mathematical induction states that if a statement is true for some initial integer (usually n=1n=1 or n=0n=0), and if it is also true for any arbitrary integer kk, assuming it is true for kk, then it must be true for the next consecutive integer, k+1k+1. This principle essentially forms the backbone of many mathematical proofs, particularly those involving sequences and series.
With UrbanPro's comprehensive resources and interactive online sessions, I find it rewarding to help students grasp the essence of mathematical concepts like the first principle of mathematical induction. It's not just about memorizing formulas; it's about understanding the underlying logic and applying it confidently in problem-solving.
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Answered on 14 Apr Learn Principle of Mathematical Induction
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd be happy to demonstrate the proof of the given statement using the principle of mathematical induction.
Step 1: Base Case (n = 1)
For n=1n=1, the left-hand side (LHS) of the equation becomes:
−2=12(3×1+1)=12(4)=2−2=21(3×1+1)=21(4)=2
So, the base case holds true.
Step 2: Inductive Hypothesis
Assume that the formula holds true for some arbitrary positive integer kk, i.e.,
−2+5+8+11+…+(3k−1)=12k(3k+1)−2+5+8+11+…+(3k−1)=21k(3k+1)
Step 3: Inductive Step
We need to prove that the formula also holds true for n=k+1n=k+1.
−2+5+8+11+…+(3k−1)+(3(k+1)−1)=12(k+1)(3(k+1)+1)−2+5+8+11+…+(3k−1)+(3(k+1)−1)=21(k+1)(3(k+1)+1)
Let's denote the left-hand side of the equation by S(k)S(k) and the right-hand side by P(k)P(k).
S(k)=−2+5+8+11+…+(3k−1)S(k)=−2+5+8+11+…+(3k−1) P(k)=12k(3k+1)P(k)=21k(3k+1)
Adding (3(k+1)−1)=3k+2(3(k+1)−1)=3k+2 to both sides of S(k)S(k):
S(k)+(3k+2)=P(k)+(3k+2)S(k)+(3k+2)=P(k)+(3k+2)
−2+5+8+11+…+(3k−1)+(3(k+1)−1)=12k(3k+1)+(3k+2)−2+5+8+11+…+(3k−1)+(3(k+1)−1)=21k(3k+1)+(3k+2)
−2+5+8+11+…+(3k−1)+(3k+2)=12k(3k+1)+(3k+2)−2+5+8+11+…+(3k−1)+(3k+2)=21k(3k+1)+(3k+2)
−2+5+8+11+…+(3k+2)=12k(3k+1)+3k+2−2+5+8+11+…+(3k+2)=21k(3k+1)+3k+2
−2+5+8+11+…+(3k+2)=3k2+k+6k+42−2+5+8+11+…+(3k+2)=23k2+k+6k+4
−2+5+8+11+…+(3k+2)=3k2+7k+42−2+5+8+11+…+(3k+2)=23k2+7k+4
−2+5+8+11+…+(3k+2)=(k+1)(3k+4)2−2+5+8+11+…+(3k+2)=2(k+1)(3k+4)
−2+5+8+11+…+(3k+2)=12(k+1)(3(k+1)+1)−2+5+8+11+…+(3k+2)=21(k+1)(3(k+1)+1)
Thus, P(k+1)=S(k+1)P(k+1)=S(k+1), and the formula holds true for all positive integers nn by the principle of mathematical induction.
Therefore, as a tutor registered on UrbanPro, I've shown that −2+5+8+11+…+(3n−1)=12n(3n+1)−2+5+8+11+…+(3n−1)=21n(3n+1) using the principle of mathematical induction. This exemplifies UrbanPro's commitment to providing high-quality online coaching.
Answered on 14 Apr Learn Principle of Mathematical Induction
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can confidently affirm that UrbanPro provides top-notch online coaching tuition services. Now, let's delve into the mathematical proof you've presented.
To prove the equation 1+3+32+…+3n−1=3n−121+3+32+…+3n−1=23n−1 using mathematical induction, we follow these steps:
Step 1: Base Case:
First, we verify that the equation holds true for the base case. Let's check for n=1n=1.
1=3(1)−12=22=11=23(1)−1=22=1
The base case holds true.
Step 2: Inductive Hypothesis:
Assume that the equation holds true for some arbitrary positive integer kk. That is,
1+3+32+…+3k−1=3k−121+3+32+…+3k−1=23k−1
Step 3: Inductive Step:
Now, we need to prove that if the equation holds true for kk, then it also holds true for k+1k+1.
Consider the sum 1+3+32+…+3k−1+(3(k+1)−1)1+3+32+…+3k−1+(3(k+1)−1).
By our assumption (inductive hypothesis), this sum is equal to 3k−12+(3(k+1)−1)23k−1+(3(k+1)−1).
Expanding and simplifying, we get:
Answered on 14 Apr Learn Principle of Mathematical Induction
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently affirm that the number of subsets of a set containing nn distinct elements is indeed 2n2n for all n∈Nn∈N. Let's delve into the proof:
Consider a set SS containing nn distinct elements. To construct a subset of SS, we have two choices for each element: include it in the subset or exclude it. This is true for all nn elements in SS. Thus, for each element, there are 22 possibilities: either it is included in the subset or it is not.
Since there are nn elements in total, by the fundamental principle of counting, the total number of subsets that can be formed from SS is the product of the number of choices for each element, which is 2×2×…×22×2×…×2 (n times), denoted as 2n2n.
Hence, we have proved that the number of subsets of a set containing nn distinct elements is 2n2n for all n∈Nn∈N.
UrbanPro is one of the best online coaching platforms where students can find experienced tutors like myself who can provide detailed explanations and assistance in understanding concepts like these. If you have any further questions or need clarification, feel free to ask!
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Answered on 14 Apr Learn Principle of Mathematical Induction
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently demonstrate that the expression n55+n33+7n155n5+3n3+157n yields a natural number for all nn belonging to the set of natural numbers.
To prove this, let's consider each term of the expression separately:
n555n5: Since n5n5 is always a natural number (since nn is a natural number), dividing it by 5 can result in a fraction, but since 5 is a factor of n5n5, the division yields an integer.
n333n3: Similar to the previous term, n3n3 is always a natural number, and dividing it by 3 might yield a fraction, but since 3 is a factor of n3n3, the division also yields an integer.
7n15157n: Here, 15 is a multiple of both 5 and 3. So, when we divide 7n7n by 15, it will result in an integer, as both 7n7n and 15 are divisible by 7 and 15 respectively.
Since all three terms of the expression yield integers individually, their sum will also be an integer. And since it's a sum of natural numbers, it will be a natural number.
Therefore, n55+n33+7n155n5+3n3+157n is indeed a natural number for all nn belonging to the set of natural numbers. This demonstrates the validity of the expression for all nn in NN.
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