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${1^4} + {2^4} + .......... + {n^4}$

Find a if ${S_n} = {1^4} + {2^4} + .......... + {n^4} = a{n^5} + b{n^4} + c{n^3} + d{n^2} + en + f$, where n is a natural number and a, b, c, d, e, f are constants.

Solution

We have, ${S_{n - 1}} = {1^4} + {2^4} + .......... + {(n - 1)^4}$

$\therefore {S_n} - {S_{n - 1}} = {n^4}$

$\therefore a\{ {n^5} - {(n - 1)^5}\}  + b\{ {n^4} - {(n - 1)^4}\}  + c\{ {n^3} - {(n - 1)^3}\}  + ..... = {n^4}$

In LHS, $n^4$ can only come from $a\{ {n^5} - {(n - 1)^5}\} $ as other terms have lower power on n.

Using binomial expansion of ${(n - 1)^5}$ and equating the coefficient of $n^4$,

$a.5 = 1$

$ \Rightarrow a = \frac{1}{5}$