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Question Number 213960 by Tawa11 last updated on 22/Nov/24

Commented by Tawa11 last updated on 22/Nov/24

$$\mathrm{Prove}\mathrm{by}\mathrm{Mathematical}\mathrm{Induction}$$

Answered by A5T last updated on 24/Nov/24

$$S_n=\underset{k=1}{\sum ^n}k(n-k)=\frac{(n-1)n(n+1)}{6}$$$$Assertiontrueforn=1,2$$$$RTP:S_n\Rightarrow S_{n+1}$$$$S_n=\underset{k=1}{\sum ^n}k(n-k)=\underset{k=1}{\sum ^n}nk-\underset{k=1}{\sum ^n}{k}^2=\frac{(n-1)n(n+1)}{6}$$$$\Rightarrow S_{n+1}=\underset{k=1}{\sum ^{n+1}}k(n+1-k)=\left(\underset{k=1}{\sum ^{n+1}}nk\right)+\underset{k=1}{\sum ^{n+1}}k-\left(\underset{k=1}{\sum ^{n+1}}{k}^2\right)$$$$=\underset{k=1}{\sum ^n}nk+n(n+1)+\frac{(n+1)(n+2)}{2}-\underset{k=1}{\sum ^n}{k}^2-{(n+1)}^2$$$$=\underset{k=1}{\sum ^n}nk-\underset{k=1}{\sum ^n}{k}^2+\frac{(n+1)\left(n\right)}{2}$$$$=\frac{(n-1)n(n+1)}{6}+\frac{3n(n+1)}{6}=\frac{(n+1)[3n+n^2-n]}{6}$$$$\Rightarrow S_{n+1}=\frac{n(n+1)(n+2)}{6}$$

Commented by Tawa11 last updated on 12/Dec/24

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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