Prove by the method of induction, for all n ∈ N. 1^2 + 2^2 + 3^2+ .. +N^2 = (n(n+1)(2n+1)/6)

Question

Prove by the method of induction, for all n ∈ N.

1² + 2² + 3²+ ....+ N² = \(\frac{n(n+1)(2n+1)}6\)

Answer 1

Let P(n) = 1² + 2² + 3² +…..+ n² = 

\(\frac{n(n+1)(2n+1)}6\) for all n ∈ N.

Step I:

Put n = 1

L.H.S. = 1² = 1

R.H.S. = \(\frac{1(1+1)[2(1)+1]}6=\frac66=1\)

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Step II: Let us assume that P(n) is true for n = k.

∴ 1² + 2² + 3² +…+ k² = \(\frac{k(k+1)(2k+1)}6\).....(i)

Step III: We have to prove that P(n) is true for n = k + 1, i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 1² + 2² + 3² +…+ n² = \(\frac{n(n+1)(2n+1)}6\)