Question Number 149024 by mathdanisur last updated on 02/Aug/21
Answered by Olaf_Thorendsen last updated on 02/Aug/21
![f_n (x) = Σ_(k=1) ^n x^k = x((1−x^n )/(1−x)) = ((x−x^(n+1) )/(1−x)) x.f′(x) = xΣ_(k=1) ^n kx^(k−1) = Σ_(k=1) ^n kx^k x.f′(x) = x[(((1−(n+1)x^n )(1−x)+(x−x^(n+1) ))/((1−x)^2 ))] x.f′(x) = ((x−(n+1)x^(n+1) +nx^(n+2) )/((1−x)^2 )) For x = (1/(2022)) : Σ_(k=1) ^n (k/(2022^k )) = (((1/(2022))−((n+1)/(2022^(n+1) ))+(n/(2022^(n+2) )))/((((2021)/(2022)))^2 )) Σ_(k=1) ^n (k/(2022^k )) = ((2022)/(2021^2 )).(1−((n+1)/(2022^n ))+(n/(2022^(n+1) )))](https://www.tinkutara.com/question/Q149070.png)
$${f}_{{n}} \left({x}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{x}^{{k}} \:=\:{x}\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}\:=\:\frac{{x}−{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}} \\ $$$${x}.{f}'\left({x}\right)\:=\:{x}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{kx}^{{k}−\mathrm{1}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{kx}^{{k}} \\ $$$${x}.{f}'\left({x}\right)\:=\:{x}\left[\frac{\left(\mathrm{1}−\left({n}+\mathrm{1}\right){x}^{{n}} \right)\left(\mathrm{1}−{x}\right)+\left({x}−{x}^{{n}+\mathrm{1}} \right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\right] \\ $$$${x}.{f}'\left({x}\right)\:=\:\frac{{x}−\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} +{nx}^{{n}+\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\mathrm{For}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2022}}\:: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}}{\mathrm{2022}^{{k}} }\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2022}}−\frac{{n}+\mathrm{1}}{\mathrm{2022}^{{n}+\mathrm{1}} }+\frac{{n}}{\mathrm{2022}^{{n}+\mathrm{2}} }}{\left(\frac{\mathrm{2021}}{\mathrm{2022}}\right)^{\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}}{\mathrm{2022}^{{k}} }\:=\:\frac{\mathrm{2022}}{\mathrm{2021}^{\mathrm{2}} }.\left(\mathrm{1}−\frac{{n}+\mathrm{1}}{\mathrm{2022}^{{n}} }+\frac{{n}}{\mathrm{2022}^{{n}+\mathrm{1}} }\right) \\ $$
Commented by mathdanisur last updated on 02/Aug/21
$${Thank}\:{You}\:{Ser} \\ $$$${Answer}:\:\frac{\mathrm{2022}}{\mathrm{2021}^{\mathrm{2}} } \\ $$
Answered by prakash jain last updated on 02/Aug/21
$$\mathrm{S}=\frac{\mathrm{1}}{\mathrm{2022}}+\frac{\mathrm{2}}{\mathrm{2022}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2022}^{\mathrm{3}} }+…+\frac{{n}}{\mathrm{2022}^{{n}} }+.. \\ $$$$\frac{\mathrm{S}}{\mathrm{2022}}=\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2022}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{2022}^{\mathrm{3}} }+…+\frac{{n}−\mathrm{1}}{\mathrm{2022}^{{n}} }+… \\ $$$${subtract} \\ $$$$\frac{\mathrm{2021}{S}}{\mathrm{2022}}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2022}^{{k}} } \\ $$$$\mathrm{blue}\:\mathrm{part}\:\mathrm{is}\:\mathrm{geometric}\:\mathrm{seres} \\ $$$$\mathrm{S}=\frac{\mathrm{2022}}{\mathrm{2021}^{\mathrm{2}} } \\ $$
Commented by mathdanisur last updated on 02/Aug/21
$${Thank}\:{You}\:{Ser} \\ $$
Commented by Olaf_Thorendsen last updated on 02/Aug/21
$$\mathrm{Excellent}\:! \\ $$