Prove by the principle of mathematical induction: 12 + 22 + 32 + �

1.	Prove by the principle of mathematical induction: 12 + 22 + 32 + … + n2 = [n(n + 1) (2n + 1)]/6
Answer» Let us considering P (n) = 1² + 2² + 3² + … + n² = [n(n + 1) (2n + 1)]/6 For, n = 1 P (1) = [1(1 + 1) (2 + 1)]/6 1 = 1 P (n) is true for n = 1 Suppose P (n) is true for n = k, therefore P (k): 1² + 2² + 3² + … + k² = [k(k + 1) (2k + 1)]/6 Let’s us check for the P (n) = k + 1, therefore P(k) = 1² + 2² + 3² + – – – – – + k² + (k + 1)² = [k + 1 (k + 2) (2k + 3)]/6 = 1² + 2² + 3² + – – – – – + k² + (k + 1)² = [k + 1 (k + 2) (2k + 3)]/6 + (k + 1)² = (k +1) [(2k² + k)/6 + (k + 1)/1] = (k +1) [2k² + k + 6k + 6]/6 = (k +1) [2k² + 7k + 6]/6 = (k +1) [2k² + 4k + 3k + 6]/6 = (k +1) [2k(k + 2) + 3(k + 2)]/6 = [(k +1) (2k + 3) (k + 2)] / 6 P (n) is true for n = k + 1 Thus, P (n) is true for all n ∈ N.

Prove by the principle of mathematical induction: 12 + 22 + 32 + … + n2 = [n(n + 1) (2n + 1)]/6

Answer»

Let us considering P (n) = 1² + 2² + 3² + … + n² = [n(n + 1) (2n + 1)]/6

For, n = 1

P (1) = [1(1 + 1) (2 + 1)]/6

1 = 1

P (n) is true for n = 1

Suppose P (n) is true for n = k, therefore

P (k): 1² + 2² + 3² + … + k² = [k(k + 1) (2k + 1)]/6

Let’s us check for the P (n) = k + 1, therefore

P(k) = 1² + 2² + 3² + – – – – – + k² + (k + 1)² = [k + 1 (k + 2) (2k + 3)]/6

= 1² + 2² + 3² + – – – – – + k² + (k + 1)²

= [k + 1 (k + 2) (2k + 3)]/6 + (k + 1)²

= (k +1) [(2k² + k)/6 + (k + 1)/1]

= (k +1) [2k² + k + 6k + 6]/6

= (k +1) [2k² + 7k + 6]/6

= (k +1) [2k² + 4k + 3k + 6]/6

= (k +1) [2k(k + 2) + 3(k + 2)]/6

= [(k +1) (2k + 3) (k + 2)] / 6

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.