Question

Find the sum of squares of first n terms.

(a) \(\frac{n(n+1)}{2}\)

(b) \((\frac{n(n+1)}{2})^3\)

(c) \(\frac{n(n+1)(2n+1)}{6}\)

(d) \((\frac{n(n+1)}{2})^2\)

The question was posed to me in unit test.

This interesting question is from Sum to n Terms of Special Series in section Sequences and Series of Mathematics – Class 11

Answer 1

Right option is (c) \(\frac{n(n+1)(2n+1)}{6}\)

Explanation: Sum of squares of first n terms = 1^2+2^2+3^2+……………+n^2

k^3–(k – 1)^3=3k^2–3k + 1

On substituting k = 1, 2, 3, ……, n and adding we get,

n^3 = 3 \(\sum_{i=0}^n k^2 – 3 \sum_{i=0}^n k + n\)

n^3 = 3 \(\sum_{i=0}^n k^2 – 3 \frac{n(n+1)}{2}\) + n

\(\sum_{i=0}^n k^2 = \frac{n(n+1)(2n+1)}{6}\).