Question

If $S_2$ and $S_4$ denotes respectively the sum of the squares and the sum of the fourth powers of first $n$ natural number, then $\frac{S_4}{S_2}=$ ________.

Answer 1

$\because S_2=1^2+2^2+3^2+......+n^2$

$\Rightarrow S_2=\frac{n(n+1)(2n+1)}{6}\cdots \left(1\right)$

And $S_4=1^4+2^4+3^4+.......+n^4=\sum n^4$

We know that,

$\because n^5-(n-1{)}^5=5n^4-10n^3-5n+1\cdots (2)$

Put $n=1,2,3,.........n$ in the equation $\left(2\right)$

$\Rightarrow 1^5-0^5={5.1}^4-{10.1}^3+{10.1}^2-5.1+1$

$\Rightarrow 2^5-1^5={5.2}^4-{10.2}^3+{10.2}^2-5.2+1$

$\Rightarrow 3^5-2^5={5.3}^4-{10.3}^3+{10.3}^2-5.3+1$

$⋮⋮⋮⋮⋮⋮⋮$

$\Rightarrow n^5-(n-1{)}^5=5n^4-10n^3+10n^2-5n+1$ + ____________________________________

$n^5-0=5.\sum n^4-10\sum n^3+10\sum n^2-5\sum n+\sum 1$

$\Rightarrow n^5=5\sum n^4-10\frac{n^2(n+1{)}^2}{4}+$
​​
​​​​​$10\frac{n(n+1)(2n+1)}{6}-5\frac{n(n+1)}{2}+n$

$\Rightarrow n^5=5\sum n^4-\frac{n(n+1)}{2}\left\{\begin{array}{c}\frac{10n(n+1)}{2}-\\ \frac{10(2n+1)}{3}+5\end{array}\right\}+n$

$\Rightarrow n^5=5\sum n^4-\frac{n(n+1)}{2}\left\{\begin{array}{c}30n^2+30n-\\ \frac{40n-20+30}{6}\end{array}\right\}+n$

$\Rightarrow n^5=5\sum n^4-\frac{5n(n+1)(3n^2-n+1)}{6}+n$

$\Rightarrow 5\sum n^4=n^5+\frac{5n(n+1)(3n^2-n+1)-6n}{6}$

$\Rightarrow 5\sum n^4=\frac{6n^5+5n(n+1)(3n^2-n+1)-6n}{6}$

$\Rightarrow 5\sum n^4=\frac{6n^5+15n^4+10n^3-n}{6}$

$\Rightarrow 5\sum n^4=\frac{n}{6}(6n^4+15n^3+10n^2-1)$

$\Rightarrow 5\sum n^4=\frac{n}{6}(n+1)(2n+1)(3n^2+3n-1)$

$\Rightarrow \sum n^4=\frac{n}{30}(n+1)(2n+1)(3n^2+3n-1)$

$\therefore S_4=\frac{n}{30}(n+1)(2n+1)(3n^2+3n-1)$

$\therefore \frac{S_4}{S_2}=\frac{\frac{n}{30}(n+1)(2n+1)(3n^2+3n-1)}{\frac{n(n+1)(2n+1)}{6}}$

$\therefore \frac{S_4}{S_2}=\frac{1}{5}(3n^2+3n-1)$