<s>[INST] 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 $t_n=n^2(2n-1)$
 [/INST] Hint:
$$\sum_{i=1}^{n}(2i^3-i^2)=2\sum_{i=1}^{n}i^3-\sum_{i=1}^{n}i^2$$

$$\sum_{i=1}^{n}i^3=\frac{n^2(n+1)^2}{4}$$$$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$$
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