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Question Number 121472 by ATHISHHUZAIN last updated on 08/Nov/20

$$\mathit{F}\mathit{i}\mathit{n}\mathit{d}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{u}\mathit{m}\mathit{t}\mathit{o}\mathit{n}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{s}\mathit{o}\mathit{f}\mathit{t}\mathit{h}\mathit{e}$$$$\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}3\times 8+6\times 11+9\times 14+.......$$$$\mathit{A}nswer:$$$${\mathit{n}}^{\mathit{t}\mathit{h}}\mathit{t}\mathit{e}\mathit{r}\mathit{m}{\mathit{t}}_{\mathit{n}}=3\mathit{n}(3\mathit{n}+5)$$$$=9{\mathit{n}}^2+15\mathit{n}$$$$$$

Answered by TANMAY PANACEA last updated on 08/Nov/20

$$T_n=3n[8+(n-1)3]=3n(3n+5)=9n^2+15n$$$$S_n=\underset{n=1}{\sum ^n}{T}_n=9\underset{n=1}{\sum ^n}{n}^2+15\underset{n=1}{\sum ^n}n$$$$S_n=9\times \frac{n(n+1)(2n+1)}{6}+15\times \frac{n(n+1)}{2}$$$$$$

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