Question

What is the difference between the sum of the cubes and that of squares of the first ten natural numbers ?

• A. $5,280$
• B. $2,640$
• C. $3,820$
• D. $4,130$

Answer 1

Answer: (B)

$2,640$

Difference between sum of cubes and that of squares of $n$ natural numbers $=\sum n^3-\sum n^2={\left[\frac{n(n+1)}{2}\right]}^2-\frac{n(n+1)(2n+1)}{6}$
$=\frac{n(n+1)}{2}[\frac{n(n+1)}{2}-\frac{2n+1}{3}]$
$=\frac{n(n+1)}{2}\left[\frac{3n^2+3n-4n-2}{6}\right]$
$=\frac{n(n+1)}{2}\left[\frac{3n^2-n-2}{6}\right]$
$=\frac{n(n+1)}{12}[{3n}^2-3n+2n-2]$$=\frac{n(n+1)}{12}[3n(n-1)+2(n-1)]$
$=\frac{n(n+1)(n-1)(3n+2)}{12}$
Here we have $n=10$
Therefore, we get $\frac{10}{12}(10+1)(10-1)(3\times 10+2)$
$=\frac{10\times 11\times 9\times 32}{12}$
$=2640$