Prove by principle of mathematical induction that for all n ∈ N.12+2

1.	Prove by principle of mathematical induction that for all n ∈ N.12+22 +33+ ⋯ + n2 = \(\frac{n(n+1)n(2+1)}{6}\)
Answer» Let, P(n) ∶ 1²+2²+3³+⋯+n² = \(\frac{n(n+1)n(2+1)}{6}\) ...(i) Step 1: Put n = 1 in ..(i) P(1) ∶ 1² = \(\frac{1.(1+1)(2.1+1)}{6}\) or 1 = 1,Which is true Thus, P(1) is true. Step 2: Assume that P(K) is true for some natural number k, i.e., let P(k) ∶ 1²+2²+3³+⋯+k² = \(\frac{k(k+1)(2k+1)}{6}\) ...(ii) Step 3: To prove that P(k+1) ∶ 1²+2²+3³+⋯+k²+(k + 1)² \(=\frac{(k+1)(k+2)(2k+3)}{6}\)is also true. Now, LHS of P(k + 1) = 1²+2²+3³+⋯+k²+(k + 1)² \(=\frac{k(k+1)(2k+1)}{6}+(k+1)\) = \(\frac{1}{6}\)(k + 1)(2k + 1) + 6(k + 1) = \(\frac{1}{6}\)(k + 1)(2k² + 7k + 6) = \(\frac{1}{6}\)(k + 1)(2k² + 4k + 3k + 6) = \(\frac{1}{6}\)(k + 1)[(2k(k + 2) + 3(k + 2))] = \(\frac{1}{6}\)(k + 1)(k + 2)(2k + 3) = RHS OF P(k + 1) ⟹ P(k + 1) is true whenever P(k) is true Hence, by principle of mathematical induction P(n) is true for all n ∈ M.

Prove by principle of mathematical induction that for all n ∈ N.12+22 +33+ ⋯ + n2 = \(\frac{n(n+1)n(2+1)}{6}\)

Answer»

Let,

P(n) ∶ 1²+2²+3³+⋯+n² = \(\frac{n(n+1)n(2+1)}{6}\) ...(i)

Step 1: Put n = 1 in ..(i)

P(1) ∶ 1² = \(\frac{1.(1+1)(2.1+1)}{6}\) or 1 = 1,Which is true Thus, P(1) is true.

Step 2: Assume that P(K) is true for some natural number k, i.e., let

P(k) ∶ 1²+2²+3³+⋯+k² = \(\frac{k(k+1)(2k+1)}{6}\) ...(ii)

Step 3: To prove that

P(k+1) ∶ 1²+2²+3³+⋯+k²+(k + 1)²
\(=\frac{(k+1)(k+2)(2k+3)}{6}\)is also true.

Now, LHS of P(k + 1)

= 1²+2²+3³+⋯+k²+(k + 1)²

\(=\frac{k(k+1)(2k+1)}{6}+(k+1)\)

= \(\frac{1}{6}\)(k + 1)(2k + 1) + 6(k + 1)

= \(\frac{1}{6}\)(k + 1)(2k² + 7k + 6)

= \(\frac{1}{6}\)(k + 1)(2k² + 4k + 3k + 6)

= \(\frac{1}{6}\)(k + 1)[(2k(k + 2) + 3(k + 2))]

= \(\frac{1}{6}\)(k + 1)(k + 2)(2k + 3) = RHS OF P(k + 1)

⟹ P(k + 1) is true whenever P(k) is true Hence, by principle of mathematical induction P(n) is true for all n ∈ M.