Question

Find the sum to $n$ terms of the series: ${1.2}^2+{3.3}^2+{5.4}^2+{7.5}^2+.....$

Answer 1

The general term for the given series is $=\sum _{k=2}^{k=n}(2k-3)\left(k^2\right)$

Now

$\sum _{k=2}^{k=n}(2k-3)\left(k^2\right)=2\sum _{k=2}^{k=n}{k}^3-3\sum _{k=2}^{k=n}{k}^2$

Now we know that

$\sum _{k=1}^{k=n}{k}^2=\frac{n(n+1)(2n+1)}{6}$ and

$\sum _{k=1}^{k=n}{k}^3=\frac{n^2(n+1{)}^2}{4}$

$\sum _{k=2}^{k=n}(2k-3)\left(k^2\right)=$ $2(\frac{n^2(n+1{)}^2}{4}-1)$ $-3(\frac{n(n+1)(2n+1)}{6}-1)$

$\sum _{k=2}^{k=n}(2k-3)\left(k^2\right)=$ $\frac{n^2(n+1{)}^2}{2}-\frac{n(n+1)(2n+1)}{2}+1$