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Question Number 106730 by I want to learn more last updated on 06/Aug/20

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$$\underset{n=1}{\sum ^n}\frac{n^2{(n+1)}^2}{4n^2}=\frac{1}{4}(\sum ^n{n}^2+\sum ^{n}2n+\sum ^{n}1)$$$$=\frac{1}{4}(\frac{n(n+1)(2n+1)}{6}+n(n+1)+n)$$$$=\frac{1}{4}((n+1)(\frac{2n^2+1}{6}+n)+n)$$$$=\frac{1}{4}(\frac{1}{6}(n+1)(2n^2+6n+1)+n)$$$$=\frac{1}{24}.17.609+4$$$$$$

Commented by I want to learn more last updated on 06/Aug/20

$$\mathrm{Thanks}\mathrm{sir}.$$

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