Question-216367

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))

$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }=? \\ $$$${a}_{{n}} =\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)−{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2021}} {\sum}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$ \\ $$$$\:\:\:=\left(\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\cancel{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:\:\:\:+\left(\cancel{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }}−\cancel{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:\:\:+\left(\:\cancel{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }}−\cancel{\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:\:\:… \\ $$$$\:\:\:\:\:\:+\left(\cancel{\frac{\mathrm{1}}{\left(\mathrm{2021}\right)^{\mathrm{2}} }}−\frac{\mathrm{1}}{\left(\mathrm{2022}\right)^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2022}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\frac{\left(\mathrm{2022}\right)^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{2022}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\frac{\left(\mathrm{2022}−\mathrm{1}\right)\left(\mathrm{2022}+\mathrm{1}\right)}{\left(\mathrm{2022}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\frac{\left(\mathrm{2021}\right)\left(\mathrm{2023}\right)}{\left(\mathrm{2022}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{OR} \\ $$$$\:\:\:\:=\frac{\mathrm{4088483}}{\mathrm{4088484}} \\ $$

Commented by Ismoiljon_008 last updated on 08/Feb/25

$$\:\:\:{thank}\:{you}\:{very}\:{much} \\ $$

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