Question

Find the sum of the cubes of first five natural numbers. Verify your answer by using formula $\left(n\right)(n+1)/2{]}^2$, where $n$ is a natural numbers. Also find the sum of the cubs of the first ten natural numbers.

Answer 1

$S=1^3+2^3+3^3+4^3............+n^3$

$n^4-(n-1{)}^1=4n^3-6n^2+4n-1$

$n^4=4S-\frac{6n(n+1)(2n+1)}{6}+2n(n+1)-n$

$4S=n^4+n(n+1)(2n+1)-2n(n+1)+n$

$4S=n^4+n(2n^2+3n+1)-2n^2-2n+n$

$4A=n^4+2n^3+3n^2+n-2n^2-2n+n$

$4S=n^4+2n^3+n^2$

$4S=n^2(n^2+2n+1)$

$4S=n^2(n+1{)}^2$

$S={\left[\frac{n(n+1)}{2}\right]}^2$

first ten terms

$S={\left[\frac{10\left(11\right)}{2}\right]}^2$

$S=55\times 55$

$=3025$

$5555\_\_\_275275\times \_\_\_\_\_\_3025$