Question-170536

Question Number 170536 by daus last updated on 26/May/22

Answered by naka3546 last updated on 26/May/22

$$\mathrm{N}\:=\:\mathrm{99} \\ $$

Answered by floor(10²Eta[1]) last updated on 26/May/22

((2k+1)/(k^2 (k+1)^2 ))=(A/k)+(B/k^2 )+(C/(k+1))+(D/((k+1)^2 )) 2k+1=Ak(k+1)^2 +B(k+1)^2 +Ck^2 (k+1)+Dk^2 (A+C)k^3 +(2A+B+C+D)k^2 +(A+2B)k+B A=−C 2A+B+C+D=0 A+2B=2 B=1 ⇒A=0=C, D=−1 ((2k+1)/((k^2 +k)^2 ))=(1/k^2 )−(1/((k+1)^2 )) ⇒Σ_(k=1) ^N (1/k^2 )−(1/((k+1)^2 ))=(1/1)−(1/2^2 )+(1/2^2 )−(1/3^2 )+(1/3^2 )−(1/4^2 )+...+(1/((N−1)^2 ))−(1/N^2 )+(1/N^2 )−(1/((N+1)^2 )) =1−(1/((N+1)^2 ))=((9999)/(10000)) (1/((N+1)^2 ))=(1/(10000))⇒10^4 =(N+1)^2 ⇒N+1=100⇒N=99

$$\frac{\mathrm{2k}+\mathrm{1}}{\mathrm{k}^{\mathrm{2}} \left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{A}}{\mathrm{k}}+\frac{\mathrm{B}}{\mathrm{k}^{\mathrm{2}} }+\frac{\mathrm{C}}{\mathrm{k}+\mathrm{1}}+\frac{\mathrm{D}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2k}+\mathrm{1}=\mathrm{Ak}\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{B}\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{Ck}^{\mathrm{2}} \left(\mathrm{k}+\mathrm{1}\right)+\mathrm{Dk}^{\mathrm{2}} \\ $$$$\left(\mathrm{A}+\mathrm{C}\right)\mathrm{k}^{\mathrm{3}} +\left(\mathrm{2A}+\mathrm{B}+\mathrm{C}+\mathrm{D}\right)\mathrm{k}^{\mathrm{2}} +\left(\mathrm{A}+\mathrm{2B}\right)\mathrm{k}+\mathrm{B} \\ $$$$\mathrm{A}=−\mathrm{C} \\ $$$$\mathrm{2A}+\mathrm{B}+\mathrm{C}+\mathrm{D}=\mathrm{0} \\ $$$$\mathrm{A}+\mathrm{2B}=\mathrm{2} \\ $$$$\mathrm{B}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{A}=\mathrm{0}=\mathrm{C},\:\mathrm{D}=−\mathrm{1} \\ $$$$\frac{\mathrm{2k}+\mathrm{1}}{\left(\mathrm{k}^{\mathrm{2}} +\mathrm{k}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\left(\mathrm{N}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{N}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{N}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{N}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{N}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{9999}}{\mathrm{10000}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{N}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{10000}}\Rightarrow\mathrm{10}^{\mathrm{4}} =\left(\mathrm{N}+\mathrm{1}\right)^{\mathrm{2}} \Rightarrow\mathrm{N}+\mathrm{1}=\mathrm{100}\Rightarrow\mathrm{N}=\mathrm{99} \\ $$

Commented by naka3546 last updated on 26/May/22

$$\mathrm{Nice}\:\:\mathrm{solution}.. \\ $$

Commented by SLVR last updated on 27/May/22

$${great}\:…{sir} \\ $$

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