Question

$1^4+2^4+3^4+\dots +n^4=$

• A. $\frac{n(n+1)(2n+1)}{30}$
• B. $\frac{n(n+1)(2n+1)(3n^2+3n-1)}{6}$
• C. $\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$
• D. $\frac{n(n+1)(2n+1{)}^2}{30}$

Answer 1

Answer: (C)

$\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$

$S\left(n\right)=1^4+2^4+3^4+\cdots +n^4$
$\therefore$ $S\left(n\right)$ will be a fifth degree polynomial of the form.

$S\left(n\right)=a{n}^5+b{n}^4+c{n}^3+d{n}^2+en+f$
$\because S\left(0\right)=0\Rightarrow f=0$

$\because S(n-1)=a{(n-1)}^5+b{(n-1)}^4+c{(n-1)}^3+d{(n-1)}^2+e(n-1)$
$\because S\left(n\right)-S(n-1)=\left(5a\right)n^4+(-10a+4b)n^3+(10a-6b+3c)n^2+(-5a+4b-3c+2d)n+(a-b+c-d+e)$

$\because$ This has to be equal to $n^4$,
$\therefore 5a=1-10a+4b=010a-6b+3c=0-5a+4b-3c+2d=0a-b+c-d+e=0$
$\Rightarrow a=\frac{1}{5},b=\frac{1}{2},c=\frac{1}{3},d=0,e=\frac{-1}{30}$
$\Rightarrow S\left(n\right)=\frac{1}{5}{n}^5+\frac{1}{2}{n}^4+\frac{1}{3}{n}^3-\frac{1}{30}n$

By simplification we get,

$\Rightarrow S\left(n\right)=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$