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Question-40743

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Question Number 40743 by scientist last updated on 27/Jul/18
Answered by maxmathsup by imad last updated on 27/Jul/18
we have Σ_(k=0) ^∞ x^k =(1/(1−x)) for ∣x∣<1 ⇒Σ_(k=1) ^∞ k x^(k−1) =(1/((1−x)^2 )) ⇒((1+x)/((1−x)^2 )) = (1+x)Σ_(k=1) ^∞ k x^(k−1) =Σ_(k=1) ^∞ k x^(k−1) +Σ_(k=1) ^∞ kx^k so the coefficient of x^(n−1) is λ =n +n−1 =2n−1 we have ((1+x)/((1−x)^2 )) = Σ_(k=0) ^∞ (k+1)x^k +Σ_(k=0) ^∞ kx^k =Σ_(k =0) ^∞ (2k+1)x^k = Σ_(k=0) ^(n−1) (2k+1)x^k +Σ_(k=n) ^∞ (2k+1)x^k ⇒ Σ_(k=n) ^∞ (2k+1)x^k =((1+x)/((1−x)^2 )) −Σ_(k=0) ^(n−1) (2k+1)x^k but Σ_(k=0) ^(n−1) (2k+1)x^k =2Σ_(k=0) ^(n−1) kx^k +Σ_(k=0) ^(n−1) x^k Σ_(k=0) ^(n−1) x^k =((1−x^n )/(1−x)) also we have Σ_(k=0) ^N x^k =((x^(N+1) −1)/(x−1)) ⇒ Σ_(k=1) ^N kx^(k−1) =((Nx^(N+1) −(N+1)x^N +1)/((1−x)^2 )) ⇒ Σ_(k=0) ^(n−1) kx^k =(x/((1−x)^2 )){(n−1)x^n −nx^(n−1) +1) ⇒ Σ_(k=n) ^∞ (2k+1)x^k =((1+x)/((1−x)^2 )) −((2x)/((1−x)^2 )){(n−1)x^n −nx^(n−1) +1}−((1−x^n )/(1−x)) ...
$${we}\:{have}\:\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{x}^{{k}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:{for}\:\:\mid{x}\mid<\mathrm{1}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} {k}\:{x}^{{k}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:=\:\left(\mathrm{1}+{x}\right)\sum_{{k}=\mathrm{1}} ^{\infty} \:{k}\:{x}^{{k}−\mathrm{1}} \:=\sum_{{k}=\mathrm{1}} ^{\infty} \:{k}\:{x}^{{k}−\mathrm{1}} \:+\sum_{{k}=\mathrm{1}} ^{\infty} \:{kx}^{{k}} \:\:{so}\:{the}\:{coefficient} \\ $$$${of}\:{x}^{{n}−\mathrm{1}} \:{is}\:\:\lambda\:\:={n}\:+{n}−\mathrm{1}\:=\mathrm{2}{n}−\mathrm{1} \\ $$$${we}\:{have}\:\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:=\:\sum_{{k}=\mathrm{0}} ^{\infty} \left({k}+\mathrm{1}\right){x}^{{k}} \:+\sum_{{k}=\mathrm{0}} ^{\infty} \:{kx}^{{k}} \\ $$$$=\sum_{{k}\:=\mathrm{0}} ^{\infty} \left(\mathrm{2}{k}+\mathrm{1}\right){x}^{{k}} \:=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right){x}^{{k}} \:+\sum_{{k}={n}} ^{\infty} \:\left(\mathrm{2}{k}+\mathrm{1}\right){x}^{{k}} \:\Rightarrow \\ $$$$\sum_{{k}={n}} ^{\infty} \:\left(\mathrm{2}{k}+\mathrm{1}\right){x}^{{k}} \:=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right){x}^{{k}} \:\:{but} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right){x}^{{k}} \:=\mathrm{2}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{kx}^{{k}} \:+\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{x}^{{k}} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{x}^{{k}} \:=\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}\:\:{also}\:{we}\:{have}\:\:\sum_{{k}=\mathrm{0}} ^{{N}} {x}^{{k}} \:=\frac{{x}^{{N}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{N}} \:\:{kx}^{{k}−\mathrm{1}} \:=\frac{{Nx}^{{N}+\mathrm{1}} −\left({N}+\mathrm{1}\right){x}^{{N}} \:+\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} {kx}^{{k}} \:=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\left\{\left({n}−\mathrm{1}\right){x}^{{n}} −{nx}^{{n}−\mathrm{1}} +\mathrm{1}\right)\:\Rightarrow \\ $$$$\sum_{{k}={n}} ^{\infty} \left(\mathrm{2}{k}+\mathrm{1}\right){x}^{{k}} \:=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:−\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\left\{\left({n}−\mathrm{1}\right){x}^{{n}} −{nx}^{{n}−\mathrm{1}} +\mathrm{1}\right\}−\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}\:… \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18
((1+x)/((1−x)^2 ))=(1+x)(1−x)^(−2) =(1+2x+3x^2 +4x^3 +...(r+1)x^r +....)(1+x) so the terms containing x^(n−1) are (n−1+1)x^(n−1) +x.(1+n−2)x^(n−2) =n.x^(n−1) +(n−1)x^(n−1) =(2n−1)x^(n−1) hence the coefficient of x^(n−1) is2n−1
$$\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)^{−\mathrm{2}} \\ $$$$\:\:=\left(\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +…\left({r}+\mathrm{1}\right){x}^{{r}} +….\right)\left(\mathrm{1}+{x}\right) \\ $$$${so}\:{the}\:{terms}\:{containing}\:{x}^{{n}−\mathrm{1}} \:\:{are} \\ $$$$\left({n}−\mathrm{1}+\mathrm{1}\right){x}^{{n}−\mathrm{1}} +{x}.\left(\mathrm{1}+{n}−\mathrm{2}\right){x}^{{n}−\mathrm{2}} \\ $$$$={n}.{x}^{{n}−\mathrm{1}} +\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{1}} \\ $$$$=\left(\mathrm{2}{n}−\mathrm{1}\right){x}^{{n}−\mathrm{1}} \\ $$$${hence}\:{the}\:{coefficient}\:{of}\:{x}^{{n}−\mathrm{1}} \:{is}\mathrm{2}{n}−\mathrm{1} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18
(1+2x+3x^2 +4x^3 +...(r+1)x^r +...)(1+x) ={1+x(1+2)+x^2 (2+3)+x^3 (3+4)+..x^r (r+r+1) .... =1+3x+5x^2 +7x^3 +...(2r+1)x^r +.... so nth term is {1+(n−1)2}x^(n−1) (2n−1)x^(2n−1)
$$\left(\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +…\left({r}+\mathrm{1}\right){x}^{{r}} +…\right)\left(\mathrm{1}+{x}\right) \\ $$$$=\left\{\mathrm{1}+{x}\left(\mathrm{1}+\mathrm{2}\right)+{x}^{\mathrm{2}} \left(\mathrm{2}+\mathrm{3}\right)+{x}^{\mathrm{3}} \left(\mathrm{3}+\mathrm{4}\right)+..{x}^{{r}} \left({r}+{r}+\mathrm{1}\right)\right. \\ $$$$…. \\ $$$$=\mathrm{1}+\mathrm{3}{x}+\mathrm{5}{x}^{\mathrm{2}} +\mathrm{7}{x}^{\mathrm{3}} +…\left(\mathrm{2}{r}+\mathrm{1}\right){x}^{{r}} +…. \\ $$$${so}\:{nth}\:{term}\:{is}\:\left\{\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{2}\right\}{x}^{{n}−\mathrm{1}} \\ $$$$\left(\mathrm{2}{n}−\mathrm{1}\right){x}^{\mathrm{2}{n}−\mathrm{1}} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18
T_(n+1) ={2n+2−1}x^(2n+2−1) =(2n+1)x^(2n+1) T_(n+2) ={2n+4−1}x^(2n+4−1) =(2n+3)x^(2n+3) T_(n+3) ={2n+6−1}x^(2n+6−1) =(2n+5)x^(2n+5) ..... .... S=T_(n+1) +T_(n+2) +T_(n+3) +.....upto ∞ =2n(x^(2n+1) +x^(2n+3) +x^(2n+5) +...)+(1.x^(2n+1) +3x^(2n+3) 5.x^(2n+5) +...) =2n.x^(2n+1) ((1/(1−x^2 )))+(1.x^(2n+1) +3.x^(2n+3) +5.x^(2n+5) ...) let S_k =1.x^(2n+1) +3.x^(2n+3) +5.x^(2n+5) +... x^2 S_k = +1.x^(2n+3) +3.x^(2n+5) +.... S_k (1−x^2 )=1.x^(2n+1) +2(x^(2n+3) +x^(2n+5) +....) S_k (1−x^2 )=1.x^(2n+1) +2.(x^(2n+3) /(1−x^2 )) S_k =(x^(2n+1) /(1−x^2 ))+((2.x^(2n+3) )/((1−x^2 )^2 )) so required sum is ((2nx^(2n+1) )/(1−x^2 ))+(x^(2n+1) /(1−x^2 ))+((2x^(2n+3) )/((1−x^2 )^2 )) =(((2n+1)x^(2n+1) )/(1−x^2 ))+((2x^(2n+3) )/((1−x^2 )^2 ))
$${T}_{{n}+\mathrm{1}} =\left\{\mathrm{2}{n}+\mathrm{2}−\mathrm{1}\right\}{x}^{\mathrm{2}{n}+\mathrm{2}−\mathrm{1}} =\left(\mathrm{2}{n}+\mathrm{1}\right){x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$${T}_{{n}+\mathrm{2}} =\left\{\mathrm{2}{n}+\mathrm{4}−\mathrm{1}\right\}{x}^{\mathrm{2}{n}+\mathrm{4}−\mathrm{1}} =\left(\mathrm{2}{n}+\mathrm{3}\right){x}^{\mathrm{2}{n}+\mathrm{3}} \\ $$$${T}_{{n}+\mathrm{3}} =\left\{\mathrm{2}{n}+\mathrm{6}−\mathrm{1}\right\}{x}^{\mathrm{2}{n}+\mathrm{6}−\mathrm{1}} =\left(\mathrm{2}{n}+\mathrm{5}\right){x}^{\mathrm{2}{n}+\mathrm{5}} \\ $$$$….. \\ $$$$…. \\ $$$${S}={T}_{{n}+\mathrm{1}} +{T}_{{n}+\mathrm{2}} +{T}_{{n}+\mathrm{3}} +…..{upto}\:\infty \\ $$$$=\mathrm{2}{n}\left({x}^{\mathrm{2}{n}+\mathrm{1}} +{x}^{\mathrm{2}{n}+\mathrm{3}} +{x}^{\mathrm{2}{n}+\mathrm{5}} +…\right)+\left(\mathrm{1}.{x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{3}{x}^{\mathrm{2}{n}+\mathrm{3}} \right. \\ $$$$\left.\:\:\:\mathrm{5}.{x}^{\mathrm{2}{n}+\mathrm{5}} +…\right) \\ $$$$=\mathrm{2}{n}.{x}^{\mathrm{2}{n}+\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)+\left(\mathrm{1}.{x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{3}.{x}^{\mathrm{2}{n}+\mathrm{3}} +\mathrm{5}.{x}^{\mathrm{2}{n}+\mathrm{5}} …\right) \\ $$$${let} \\ $$$${S}_{{k}} =\mathrm{1}.{x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{3}.{x}^{\mathrm{2}{n}+\mathrm{3}} +\mathrm{5}.{x}^{\mathrm{2}{n}+\mathrm{5}} +… \\ $$$${x}^{\mathrm{2}} {S}_{{k}} =\:\:\:\:\:\:\:\:\:\:\:+\mathrm{1}.{x}^{\mathrm{2}{n}+\mathrm{3}} +\mathrm{3}.{x}^{\mathrm{2}{n}+\mathrm{5}} +…. \\ $$$${S}_{{k}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\mathrm{1}.{x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{2}\left({x}^{\mathrm{2}{n}+\mathrm{3}} +{x}^{\mathrm{2}{n}+\mathrm{5}} +….\right) \\ $$$${S}_{{k}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\mathrm{1}.{x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{2}.\frac{{x}^{\mathrm{2}{n}+\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${S}_{{k}} =\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{2}.{x}^{\mathrm{2}{n}+\mathrm{3}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${so}\:{required}\:{sum}\:{is} \\ $$$$\frac{\mathrm{2}{nx}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{2}{x}^{\mathrm{2}{n}+\mathrm{3}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right){x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{2}{x}^{\mathrm{2}{n}+\mathrm{3}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$

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