Find the sum to $n$ terms of the series: $1^2.1+2^2.3+3^2.5+....$ Find the sum to $n$ terms of the series:
$$1^2.1+2^2.3+3^2.5+.....$$
My Attempt:
Here, $n^{th}$ term of $1,2,3,....=n$
$n^{th}$ term of $1^2,2^2,3^2,....=n^2$
Also, $n^{th}$ term of $1,3,5,....=2n-1$
Hence, $n^{th}$ term of the given series is $t_n=n^2(2n-1)$

$\displaystyle \sum_{k=1}^nk^2(2k-1)=2\sum_{k=1}^nk^3-\sum_{k=1}^nk^2=2\times\frac{1}{4}n^2(n+1)^2-\frac{1}{6}n(n+1)(2n+1)$
