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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,

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.

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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.

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