Prove by induction that the sum of the cubes of three consecutive natural numbers is divisible by 9.

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.