Question

Find the sum of $n$ terms of the series whose $n^{th}$ term is
$4n(n^2+1)-(6n^2+1)$.

Answer 1

The $n^{th}\text{term}$ term of the series is $4n(n^2+1)-(6n^2+1)=4n^3-6n^2+4n-1$

Now we need to find the summation of first $n\text{terms}$ is given by $4\sum n^3-6\sum n^2+4\sum n-\sum 1$

$4{\left(\frac{n(n+1)}{2}\right)}^2-6\frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}-n$

$=n^4+2n^3+n^2-2n^3-3n^2-n+2n^2+2n-n=n^4$