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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»

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

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


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