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Question Number 93340 by john santu last updated on 12/May/20

$$\underset{\mathrm{k}=1}{\sum ^{2011}}\frac{1}{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}?$$

Commented by mathmax by abdo last updated on 12/May/20

$$let{A}_n=\sum _{k=1}^n\frac{1}{k(k+1)(k+2)}firstletdecomposeF\left(x\right)=\frac{1}{x(x+1)(x+2)}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{x+2}\Rightarrow a=\frac{1}{2},b=-1,c=\frac{1}{2}\Rightarrow$$$$F\left(x\right)=\frac{1}{2x}-\frac{1}{x+1}+\frac{1}{2(x+2)}\Rightarrow A_n=\frac{1}{2}\sum _{k=1}^n\frac{1}{k}-\sum _{k=1}^n\frac{1}{k+1}+\frac{1}{2}\sum _{k=1}^n\frac{1}{k+2}$$$$\sum _{k=1}^n\frac{1}{k}=H_n,\sum _{k=1}^n\frac{1}{k+1}=\sum _{k=2}^{n+1}\frac{1}{k}=H_n+\frac{1}{n+1}-1$$$$\sum _{k=1}^n\frac{1}{k+2}=\sum _{k=3}^{n+2}\frac{1}{k}=H_n+\frac{1}{n+2}-\frac{3}{2}\Rightarrow$$$$A_n=\frac{1}{2}{H}_n-H_n-\frac{1}{n+1}+1+\frac{1}{2}{H}_n+\frac{1}{2(n+2)}-\frac{3}{4}\Rightarrow$$$$A_n=\frac{1}{4}+\frac{1}{2(n+2)}-\frac{1}{n+1}$$$$n=2011\Rightarrow A_{2011}=\frac{1}{4}+\frac{1}{2\times 2013}-\frac{1}{2012}$$

Commented by prakash jain last updated on 12/May/20

$$\mathrm{I}\mathrm{checked}\mathrm{my}\mathrm{answer}\mathrm{but}$$$$\mathrm{couldnt}\mathrm{see}\mathrm{any}\mathrm{obvious}\mathrm{error}.$$$$\mathrm{Request}\mathrm{your}\mathrm{review}.$$

Commented by prakash jain last updated on 12/May/20

$$abdosir$$$$\mathrm{in}\mathrm{my}\mathrm{answer}\mathrm{i}\mathrm{got}$$$$\frac{1}{4}-\frac{1}{2\cdot 2012}+\frac{1}{2\cdot 2013}$$

Commented by Ar Brandon last updated on 12/May/20

$$\mathrm{Hi}\mathrm{I}\mathrm{thought}\mathrm{it}\mathrm{was}\mathrm{gonna}\mathrm{be}$$$$\sum _{\mathrm{k}=3}^{\mathrm{n}+2}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}}+\frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2}-\frac{3}{2}$$$$\mathrm{Or}\mathrm{may}\mathrm{be}\mathrm{I}\mathrm{was}\mathrm{wrong}.$$$$\mathrm{What}\mathrm{do}\mathrm{you}\mathrm{think}?$$

Commented by turbo msup by abdo last updated on 12/May/20

$$thatswhatifinditscorrect.$$

Commented by I want to learn more last updated on 12/May/20

$${\mathrm{S}}_{\mathrm{n}}=\mathrm{C}-\frac{1}{1.2}.\frac{1}{(\mathbf{n}+1)(\mathbf{n}+2)},\mathrm{put}\mathbf{n}=1$$$$\therefore {\mathbf{S}}_1=\mathbf{C}-\frac{1}{2(1+1)(1+2)},\Rightarrow \frac{1}{1.2.3}=\mathbf{C}-\frac{1}{2\left(2\right)\left(3\right)}$$$$\therefore \frac{1}{6}=\mathbf{C}-\frac{1}{12},\mathbf{C}=\frac{1}{6}+\frac{1}{12}=\frac{2+1}{12}=\frac{3}{12}=\frac{1}{4}$$$$\therefore {\mathbf{S}}_{\mathbf{n}}=\frac{1}{4}-\frac{1}{2(\mathbf{n}+1)(\mathbf{n}+2)}$$$$\therefore {\mathbf{S}}_{2011}=\frac{1}{4}-\frac{1}{2(2011+1)(2011+2)}$$$$\therefore {\mathbf{S}}_{2011}=\frac{1}{4}-\frac{1}{2\left(2012\right)\left(2013\right)}$$$$\therefore {\mathbf{S}}_{2011}=\frac{2025077}{8100312}$$

Commented by I want to learn more last updated on 12/May/20

$$\mathrm{I}\mathrm{got}\mathrm{the}\mathrm{same}\mathrm{thing}\mathrm{as}\mathrm{prakash}\mathrm{jane}$$

Answered by prakash jain last updated on 12/May/20

$$\frac{1}{k(k+1)(k+2)}=\frac{1}{2k}-\frac{1}{(k+1)}+\frac{1}{2(k+2)}$$$$\mathrm{Sum}=$$$$\frac{1}{2\cdot 1}-\frac{1}{2}+\frac{1}{2\cdot 3}$$$$\frac{1}{2\cdot 2}-\frac{1}{3}+\frac{1}{2\cdot 4}$$$$\frac{1}{2\cdot 3}-\frac{1}{4}+\frac{1}{2\cdot 5}$$$$\frac{1}{2.4}-\frac{1}{5}+\frac{1}{2.6}$$$$..$$$$\frac{1}{2\cdot 2011}-\frac{1}{2012}+\frac{1}{2\cdot 2013}$$$$$$$$\mathrm{As}\mathrm{you}\mathrm{have}\mathrm{noticed}\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{telescopic}$$$$\mathrm{series}\left(\mathrm{middle}\mathrm{terms}\mathrm{cancel}\right).$$$$\mathrm{See}\mathrm{colored}\mathrm{terms}\mathrm{abovf}.$$$$\mathrm{final}\mathrm{sum}=\frac{1}{4}-\frac{1}{2\cdot 2012}+\frac{1}{2\cdot 2013}$$

Commented by I want to learn more last updated on 12/May/20

$$\mathrm{I}\mathrm{got}\mathrm{same}\mathrm{answer}\mathrm{as}\mathrm{you}\mathrm{sir}$$

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