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Question Number 93340 by john santu last updated on 12/May/20
Σ_(k = 1) ^(2011) (1/(k(k+1)(k+2))) ?
$$\underset{\mathrm{k}\:=\:\mathrm{1}} {\overset{\mathrm{2011}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{k}+\mathrm{2}\right)}\:? \\ $$
Commented by mathmax by abdo last updated on 12/May/20
let A_n =Σ_(k=1) ^n (1/(k(k+1)(k+2))) first let decompose F(x)=(1/(x(x+1)(x+2))) F(x)=(a/x) +(b/(x+1)) +(c/(x+2)) ⇒a =(1/2) , b =−1 ,c =(1/2) ⇒ F(x) =(1/(2x))−(1/(x+1)) +(1/(2(x+2))) ⇒A_n =(1/2)Σ_(k=1) ^n (1/k)−Σ_(k=1) ^n (1/(k+1))+(1/2)Σ_(k=1) ^n (1/(k+2)) Σ_(k=1) ^n (1/k) =H_n , Σ_(k=1) ^n (1/(k+1)) =Σ_(k=2) ^(n+1) (1/k)=H_n +(1/(n+1))−1 Σ_(k=1) ^n (1/(k+2)) =Σ_(k=3) ^(n+2) (1/k) =H_n +(1/(n+2))−(3/2) ⇒ A_n =(1/2)H_n −H_n −(1/(n+1)) +1 +(1/2)H_n +(1/(2(n+2)))−(3/4) ⇒ A_n =(1/4) +(1/(2(n+2)))−(1/(n+1)) n=2011 ⇒A_(2011) =(1/4) +(1/(2×2013))−(1/(2012))
$${let}\:{A}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\:\:{first}\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{{x}+\mathrm{2}}\:\Rightarrow{a}\:=\frac{\mathrm{1}}{\mathrm{2}}\:,\:{b}\:=−\mathrm{1}\:\:,{c}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{x}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{2}\right)}\:\Rightarrow{A}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}}\:={H}_{{n}} \:\:\:,\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}={H}_{{n}} +\frac{\mathrm{1}}{{n}+\mathrm{1}}−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{2}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} +\frac{\mathrm{1}}{{n}+\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} −{H}_{{n}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} \:+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{2}\right)}−\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$${n}=\mathrm{2011}\:\Rightarrow{A}_{\mathrm{2011}} =\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{2013}}−\frac{\mathrm{1}}{\mathrm{2012}} \\ $$
Commented by prakash jain last updated on 12/May/20
I checked my answer but couldnt see any obvious error. Request your review.
$$\mathrm{I}\:\mathrm{checked}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{but} \\ $$$$\mathrm{couldnt}\:\mathrm{see}\:\mathrm{any}\:\mathrm{obvious}\:\mathrm{error}. \\ $$$$\mathrm{Request}\:\mathrm{your}\:\mathrm{review}. \\ $$
Commented by prakash jain last updated on 12/May/20
abdo sir in my answer i got (1/4)−(1/(2∙2012))+(1/(2∙2013))
$${abdo}\:{sir} \\ $$$$\mathrm{in}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{i}\:\mathrm{got} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{2012}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{2013}} \\ $$
Commented by Ar Brandon last updated on 12/May/20
Hi I thought it was gonna be Σ_(k=3) ^(n+2) (1/k)=H_n +(1/(n+1))+(1/(n+2))−(3/2) Or may be I was wrong. What do you think?
$$\mathrm{Hi}\:\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{was}\:\mathrm{gonna}\:\mathrm{be} \\ $$$$\sum_{\mathrm{k}=\mathrm{3}} ^{\mathrm{n}+\mathrm{2}} \frac{\mathrm{1}}{\mathrm{k}}=\mathrm{H}_{\mathrm{n}} +\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{Or}\:\mathrm{may}\:\mathrm{be}\:\mathrm{I}\:\mathrm{was}\:\mathrm{wrong}.\: \\ $$$$\mathrm{What}\:\mathrm{do}\:\mathrm{you}\:\mathrm{think}? \\ $$
Commented by turbo msup by abdo last updated on 12/May/20
that s what i find its correct.
$${that}\:{s}\:{what}\:{i}\:{find}\:{its}\:{correct}. \\ $$
Commented by I want to learn more last updated on 12/May/20
S_n = C − (1/(1.2)).(1/((n + 1)(n + 2))), put n = 1 ∴ S_1 = C − (1/(2(1 + 1)(1 + 2))), ⇒ (1/(1.2.3)) = C − (1/(2(2)(3))) ∴ (1/6) = C − (1/(12)), C = (1/6) + (1/(12)) = ((2 + 1)/(12)) = (3/(12)) = (1/4) ∴ S_n = (1/4) − (1/(2(n + 1)(n + 2))) ∴ S_(2011) = (1/4) − (1/(2(2011 + 1)(2011 + 2))) ∴ S_(2011) = (1/4) − (1/(2(2012)(2013))) ∴ S_(2011) = ((2025077)/(8100312))
$$\mathrm{S}_{\mathrm{n}} \:\:=\:\:\mathrm{C}\:\:−\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}}.\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{n}}\:+\:\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}\:+\:\mathrm{2}\right)},\:\:\:\:\:\:\mathrm{put}\:\:\boldsymbol{\mathrm{n}}\:\:=\:\:\mathrm{1} \\ $$$$\therefore\:\:\:\:\boldsymbol{\mathrm{S}}_{\mathrm{1}} \:\:\:=\:\:\:\boldsymbol{\mathrm{C}}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}\:\:+\:\:\mathrm{1}\right)\left(\mathrm{1}\:\:+\:\:\mathrm{2}\right)},\:\:\:\:\Rightarrow\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:\:\:=\:\:\boldsymbol{\mathrm{C}}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\right)\left(\mathrm{3}\right)} \\ $$$$\therefore\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{6}}\:\:\:=\:\:\:\boldsymbol{\mathrm{C}}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{12}},\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{C}}\:\:\:=\:\:\:\frac{\mathrm{1}}{\mathrm{6}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{12}}\:\:\:=\:\:\:\frac{\mathrm{2}\:\:+\:\:\mathrm{1}}{\mathrm{12}}\:\:\:=\:\:\:\:\frac{\mathrm{3}}{\mathrm{12}}\:\:\:=\:\:\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{n}}} \:\:=\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{2}\left(\boldsymbol{\mathrm{n}}\:\:+\:\:\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}\:\:+\:\:\mathrm{2}\right)} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}_{\mathrm{2011}} \:\:=\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2011}\:\:+\:\:\mathrm{1}\right)\left(\mathrm{2011}\:\:+\:\:\mathrm{2}\right)} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}_{\mathrm{2011}} \:\:=\:\:\frac{\mathrm{1}}{\mathrm{4}}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2012}\right)\left(\mathrm{2013}\right)} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}_{\mathrm{2011}} \:\:=\:\:\:\frac{\mathrm{2025077}}{\mathrm{8100312}} \\ $$
Commented by I want to learn more last updated on 12/May/20
I got the same thing as prakash jane
$$\mathrm{I}\:\mathrm{got}\:\mathrm{the}\:\mathrm{same}\:\mathrm{thing}\:\mathrm{as}\:\:\mathrm{prakash}\:\mathrm{jane} \\ $$
Answered by prakash jain last updated on 12/May/20
(1/(k(k+1)(k+2)))=(1/(2k))−(1/((k+1)))+(1/(2(k+2))) Sum= (1/(2∙1))−(1/2)+(1/(2∙3)) (1/(2∙2))−(1/3)+(1/(2∙4)) (1/(2∙3))−(1/4)+(1/(2∙5)) (1/(2.4))−(1/5)+(1/(2.6)) .. (1/(2∙2011))−(1/(2012))+(1/(2∙2013)) As you have noticed this is a telescopic series (middle terms cancel). See colored terms abovf. final sum=(1/4)−(1/(2∙2012))+(1/(2∙2013))
$$\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{2}{k}}−\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({k}+\mathrm{2}\right)} \\ $$$$\mathrm{Sum}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{5}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}.\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}.\mathrm{6}} \\ $$$$.. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{2011}}−\frac{\mathrm{1}}{\mathrm{2012}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{2013}} \\ $$$$ \\ $$$$\mathrm{As}\:\mathrm{you}\:\mathrm{have}\:\mathrm{noticed}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{telescopic} \\ $$$$\mathrm{series}\:\left(\mathrm{middle}\:\mathrm{terms}\:\mathrm{cancel}\right). \\ $$$$\mathrm{See}\:\mathrm{colored}\:\mathrm{terms}\:\mathrm{abovf}. \\ $$$$\mathrm{final}\:\mathrm{sum}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{2012}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{2013}} \\ $$
Commented by I want to learn more last updated on 12/May/20
I got same answer as you sir
$$\mathrm{I}\:\mathrm{got}\:\mathrm{same}\:\mathrm{answer}\:\mathrm{as}\:\mathrm{you}\:\mathrm{sir} \\ $$

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