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

$$\sum\limits_{j=0}^n j^2(2j-1)=2\sum \limits_{j=0}^n j^3-\sum \limits_{j=0}^n j^2=2(\dfrac{n(n+1)}{2})^2-\dfrac{n(n+1)(2n+1)}{6}.$$ You can prove this formulae for $\sum j^3$ and $\sum j^2$ by induction.
