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Question Number 148427 by Rustambek last updated on 27/Jul/21

Answered by Ar Brandon last updated on 27/Jul/21

$$\mathrm{S}=2+\underset{\mathrm{k}=2}{\sum ^{2020}}\frac{(\mathrm{k}+1)\mathrm{k}}{\frac{1}{\mathrm{k}!}+\frac{1}{(\mathrm{k}-1)!}}$$$$=2+\underset{\mathrm{k}=2}{\sum ^{2020}}\frac{\mathrm{k}(\mathrm{k}+1)!}{1+\mathrm{k}}=2+\underset{\mathrm{k}=2}{\sum ^{2020}}\mathrm{k}(\mathrm{k}!)$$$$=2+\underset{\mathrm{k}=2}{\sum ^{2020}}[(\mathrm{k}+1)-1](\mathrm{k}!)=2+\underset{\mathrm{k}=2}{\sum ^{2020}}[(\mathrm{k}+1)!-\mathrm{k}!]$$$$=2+[(3!-2!)+(4!-3!)+\cdot \cdot \cdot (2021!-2020!)]$$$$=2021!$$

Commented by Rustambek last updated on 27/Jul/21

$$?$$

Answered by Olaf_Thorendsen last updated on 27/Jul/21

$$\mathrm{S}=2+\underset{k=1}{\sum ^{2019}}\frac{(k+1)(k+2)}{\frac{1}{k!}+\frac{1}{(k+1)!}}$$$$\mathrm{S}=2+\underset{k=1}{\sum ^{2019}}\frac{(k+1)(k+2)}{\frac{k+1}{(k+1)!}+\frac{1}{(k+1)!}}$$$$\mathrm{S}=2+\underset{k=1}{\sum ^{2019}}\frac{(k+1)(k+2)}{\frac{k+2}{(k+1)!}}$$$$\mathrm{S}=2+\underset{k=1}{\sum ^{2019}}(k+1)(k+1)!2+\underset{k=2}{\sum ^{2020}}k.k!$$$$\mathrm{S}=2+\underset{k=2}{\sum ^{2020}}((k+1)!-k!)$$$$$$$$\mathrm{S}=2021!$$

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