Question

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

Answer 1

R.H.S. = $\frac{(-1{)}^{n-1}}{2}n(n^5+3n^4-5n^2+3)$
$=\frac{(-1{)}^{n-1}}{2}n(n+1)(n^4+2n^3-2n^2-3n+3).....\left(1\right)$
by actual division.
P(1) is obiviously true. Assume P(n).
$\therefore$ P(n + 1) = P(n) +$(-1{)}^n(n+1{)}^6$
$=\frac{(-1{)}^{n-1}}{2}n(n+1)(n^4+2n^3-2n^2-3n+3)+(-1{)}^n(n+1{)}^6$
$=\frac{(-1{)}^n}{2}(n+1)[-(n^5+2n^4-2n^3-3n^2+3n)+2(n+1{)}^5]$
$=\frac{(-1{)}^n}{2}(n+1)\left[2\right(n^5+5n^4+10n^3+10n^2+5n+1)-n^5-2n^4+2n^3+3n^2-3n]$
$=\frac{(-1{)}^n}{2}(n+1)[n^5+8n^4+22n^3+23n^2+7n+2]$
$=\frac{(-1{)}^n}{2}(n+1)(n+2)$
$[n^4+6n^3+10n^2+3n+1]$ ....(2)
Now keeping in view the form (1) the expression within the bracket can be written as
{$(n+1{)}^4+2(n+1{)}^3-2(n+1{)}^2-3(n+1)+3$}
Hence (2) takes the form of (1) with
$n\to (n+1)\therefore P\left(n\right)\Rightarrow P(n+1)$