Question

Sum of cubes of first n natural numbers is __________
1. \(\left [\rm\frac{n(n + 1)(n+2)}{2} \right]\)
2. \(\left [\rm\frac{2n}{(n+1)} \right]^2\)
3. \( \dfrac{n(n + 1)(2n + 1)}{6}\)
4. \(\left [\rm\frac{n(n + 1)}{2} \right]^2\)

Answer 1

Correct Answer - Option 4 : \(\left [\rm\frac{n(n + 1)}{2} \right]^2\)

Sum of cubes of first n natural numbers = \(\rm \sum_{1}^{n} r^3 = \left [\rm\frac{n(n + 1)}{2} \right]^2\)

Sum of first 'n' natural numbers =  \(\rm \sum_{1}^{n} r = \dfrac{n(n + 1)}{2}\)

Sum of the square of first n natural numbers = \(\rm \sum_{1}^{n} r^2= \dfrac{n(n + 1)(2n + 1)}{6}\)