Hello, Harshal!

How about an inductive proof?

Prove: .1² + 2² + 3² + ... + n² .= .n(n + 1)(2n + 1)/6

Verify S(1): .1² .= .1·2·3/6 . → . 1 = 1 ... true

Assume S(k): .1² + 2² + 3² + ... k² .= .k(k + 1)(2k + 1)/6

Add (k + 1)² to both sides:

. . 1² + 2² + 3² + ... + (k + 1)² .= .k(k + 1)(2k + 1)/6 + (k + 1)²

The left side is the left side of S(k + 1).

Factor the right side: .(k + 1) [k(k + 1)/6 + (k + 1)]

. . = .(k + 1) (k² + k + 6k + 6)/6 .= .(k + 1)(k² + 7k + 6)/6

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

. . and this is the right side of S(k + 1).

The inductive proof is complete.