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

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  • $\begingroup$ You can then use the formulas of $\sum n^3, \sum n^2$ $\endgroup$ – King Tut Mar 7 '18 at 14:06
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$\displaystyle \sum_{k=1}^nk^2(2k-1)=2\sum_{k=1}^nk^3-\sum_{k=1}^nk^2=2\times\frac{1}{4}n^2(n+1)^2-\frac{1}{6}n(n+1)(2n+1)$

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

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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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$$\sum_{k=1}^nk^2(2k-1)=\sum_{k=1}^n\left[12\binom{k}3+10\binom{k}2+\binom{k}1\right]=12\binom{n+1}4+10\binom{n+1}3+\binom{n+1}2$$

This under the convention that $\binom{k}{r}=0$ if $r\notin\{0,\dots,k\}$.

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