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Question-216367

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Question Number 216367 by Ismoiljon_008 last updated on 05/Feb/25
Answered by Rasheed.Sindhi last updated on 06/Feb/25
Σ_(n=1) ^(2021) ((2n+1)/(n^2 (n+1)^2 ))=? a_n =((2n+1)/(n^2 (n+1)^2 )) =(((n^2 +2n+1)−n^2 )/(n^2 (n+1)^2 )) =(((n+1)^2 )/(n^2 (n+1)^2 ))−(n^2 /(n^2 (n+1)^2 )) =(1/n^2 )−(1/((n+1)^2 )) Σ_(n=1) ^(2021) ((2n+1)/(n^2 (n+1)^2 )) =Σ_(n=1) ^(2021) ((1/n^2 )−(1/((n+1)^2 ))) =( (1/1^2 )−(1/2^2 )) +((1/2^2 )−(1/3^2 )) +( (1/3^2 )−(1/4^2 )) ... +((1/((2021)^2 ))−(1/((2022)^2 ))) =(1/1^2 )−(1/(2022^2 )) (((2022)^2 −1)/((2022)^2 )) =(((2022−1)(2022+1))/((2022)^2 )) =(((2021)(2023))/((2022)^2 )) OR =((4088483)/(4088484))
∑2021n=12n+1n2(n+1)2=?an=2n+1n2(n+1)2=(n2+2n+1)−n2n2(n+1)2=(n+1)2n2(n+1)2−n2n2(n+1)2=1n2−1(n+1)2∑2021n=12n+1n2(n+1)2=∑2021n=1(1n2−1(n+1)2)=(112−122)+(122−132)+(132−142)…+(1(2021)2−1(2022)2)=112−120222(2022)2−1(2022)2=(2022−1)(2022+1)(2022)2=(2021)(2023)(2022)2OR=40884834088484
Commented by Ismoiljon_008 last updated on 08/Feb/25
thank you very much
thankyouverymuch

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