1.

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»

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

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


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