Question

Find the sum of cubes 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) \((\frac{n(n+1)}{2})^2\)

I got this question in an online quiz.

Origin of the question is Sum to n Terms of Special Series in section Sequences and Series of Mathematics – Class 11

Answer 1

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

Best explanation: Sum of cubes of first n terms = 1^3+2^3+3^3+……………+n^3

(k + 1)^4–k^4 = 4k^3 + 6k^2 + 4k + 1.

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

4n^3+n^4+6n^2+4n = \(4\sum_{i=0}^n k^3 + 6 \sum_{i=0}^n k^2 + 4 \sum_{i=0}^n k + n\)

4n^3+n^4+6n^2+4n = \(4\sum_{i=0}^n k^3 + 6\frac{(n(n+1)(2n+1))}{6} + 4\frac{n(n+1)}{2} + n\)

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