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 $t_n=n^2(2n-1)$

As , $$T_n=n^2(2n-1)$$ Hence , $$S_n=\sum T_n$$ Or , $$S_n=\sum_{k=1}^{n}2k^3-k^2$$ $$S_n=2\sum_{k=1}^{n}k^3-\sum_{k=1}^{n}k^2$$ Hence ,

$$S_n=\frac{(n(n+1))^2}{2}-\frac{n(n+1)(2n+1)}{6}$$

