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Question Number 119055 by mathocean1 last updated on 21/Oct/20

$$$$$$Showbyrecurencethat:\mathrm{\forall }n\in {\mathbb{N}}^{\ast }$$$$\sum _{k=1}^n\frac{1}{k(k+1)(k+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$$

Answered by Bird last updated on 21/Oct/20

$$n=1\frac{1}{2\times 3}=\frac{4}{4\left(2\right)\times 4}\left(rdsulttrue\right)$$$$letsupoose\sum _{k=1}^n(...)=\frac{n(n+3)}{4(n+1)(n+2)}$$$$\sum _{k=1}^{n+1}\frac{1}{k(k+1)(k+2)}$$$$=\sum _{k=1}^n\frac{1}{k(k+1)(k+2)}+\frac{1}{(n+1)(n+2)(n+3)}$$$$=\frac{n(n+3)}{4(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}$$$$=\frac{1}{(n+1)(n+2)}(\frac{n(n+3)}{4}+\frac{1}{n+3})$$$$=\frac{1}{)n+1)(n+2)}\left(\frac{n{(n+3)}^2+4}{4(n+3)}\right)$$$$=\frac{n{(n+3)}^2+4}{4(n+1)(n+2)(n+3)}$$$$isthisequalto\frac{(n+1)(n+4)}{4(n+2)(n+3)}?\mathrm{n}\Rightarrow$$$$\frac{n{(n+3)}^2+4}{n+1}=(n+1)(n+4)\Rightarrow$$$$n{(n+3)}^2+4={(\mathrm{n}+1)}^2(\mathrm{n}+4)$$$$\mathrm{let}\mathrm{prove}\mathrm{this}\mathrm{we}\mathrm{have}$$$$\mathrm{n}{(\mathrm{n}+3)}^2+4=\mathrm{n}({\mathrm{n}}^2+6\mathrm{n}+9)+4$$$$={\mathrm{n}}^3+{6\mathrm{n}}^2+9\mathrm{n}+4$$$${(\mathrm{n}+1)}^2(\mathrm{n}+4)=({\mathrm{n}}^2+2\mathrm{n}+1)(\mathrm{n}+4)$$$$={\mathrm{n}}^3+{4\mathrm{n}}^2+{2\mathrm{n}}^2+8\mathrm{n}+\mathrm{n}+4$$$$={\mathrm{n}}^3+{6\mathrm{n}}^2+9\mathrm{n}+4\mathrm{so}\mathrm{the}$$$$\mathrm{equality}\mathrm{is}\mathrm{true}\mathrm{at}\mathrm{term}\mathrm{n}+1$$

Answered by Dwaipayan Shikari last updated on 22/Oct/20

$$WithoutInduction$$$$\underset{k=1}{\sum ^n}\frac{1}{k(k+1)(k+2)}$$$$=\frac{1}{2}\underset{k=1}{\sum ^n}\frac{k+2-k}{k(k+1)(k+2)}$$$$=\frac{1}{2}\underset{k=1}{\sum ^n}\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}$$$$=\frac{1}{2}\underset{k=1}{\sum ^n}\frac{1}{k}-\frac{1}{k+1}-\frac{1}{2}\underset{k=1}{\sum ^n}\frac{1}{k+1}-\frac{1}{k+2}$$$$=\frac{1}{2}(1-\frac{1}{n+1})-\frac{1}{2}(\frac{1}{2}-\frac{1}{n+2})$$$$=\frac{2n(n+2)-n(n+1)}{4(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$$

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