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Question Number 151181 by mathdanisur last updated on 18/Aug/21
if ∣x∣<1 find x−4x^2 +9x^3 −16x^4 +...
$$\mathrm{if}\:\:\mid\boldsymbol{\mathrm{x}}\mid<\mathrm{1} \\ $$$$\mathrm{find}\:\:\mathrm{x}−\mathrm{4x}^{\mathrm{2}} +\mathrm{9x}^{\mathrm{3}} −\mathrm{16x}^{\mathrm{4}} +… \\ $$
Answered by Olaf_Thorendsen last updated on 18/Aug/21
S(x) = −Σ_(n=1) ^∞ (−1)^n n^2 x^n Let f(x) = (1/(1+x)) = Σ_(n=0) ^∞ (−1)^n x^n f′(x) = −(1/((1+x)^2 )) = Σ_(n=1) ^∞ (−1)^n nx^(n−1) xf′(x) = −(x/((1+x)^2 )) = Σ_(n=1) ^∞ (−1)^n nx^n −((1−x)/((1+x)^3 )) = Σ_(n=1) ^∞ (−1)^n n^2 x^(n−1) ((x(1−x))/((1+x)^3 )) = −Σ_(n=1) ^∞ (−1)^n n^2 x^n = S(x)
$$\mathrm{S}\left({x}\right)\:=\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {n}^{\mathrm{2}} {x}^{{n}} \\ $$$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \\ $$$${f}'\left({x}\right)\:=\:−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {nx}^{{n}−\mathrm{1}} \\ $$$${xf}'\left({x}\right)\:=\:−\frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {nx}^{{n}} \\ $$$$−\frac{\mathrm{1}−{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} \\ $$$$\frac{{x}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }\:=\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {n}^{\mathrm{2}} {x}^{{n}} \:=\:\mathrm{S}\left({x}\right) \\ $$
Commented by mathdanisur last updated on 18/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Commented by mathdanisur last updated on 19/Aug/21
Dear Ser, finally answer ((x(1-x))/((1+x)^3 )) or ((2x^2 )/((1+x)^3 )).?
$$\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{finally}\:\mathrm{answer}\:\frac{\mathrm{x}\left(\mathrm{1}-\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} } \\ $$$$\mathrm{or}\:\:\frac{\mathrm{2x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }.? \\ $$
Answered by qaz last updated on 19/Aug/21
Σ_(n=1) ^∞ n^2 (−1)^(n−1) x^n =Σ_(n=1) ^∞ (n(n+1)−n)(−1)^(n−1) x^n =Σ_(n=1) ^∞ n(n+1)(−1)^(n−1) x^n −Σ_(n=1) ^∞ n(−1)^(n−1) x^n =2Σ_(n=1) ^∞ Σ_(k=1) ^n k(−1)^(n−1) x^n −Σ_(n=1) ^∞ Σ_(k=1) ^n (−1)^(n−1) x^n =2Σ_(k=1) ^∞ Σ_(n=0) ^∞ k(−1)^(n+k−1) x^(n+k) −Σ_(k=1) ^∞ Σ_(n=0) ^∞ (−1)^(n+k−1) x^(n+k) =(2/(1+x))Σ_(k=0) ^∞ Σ_(i=0) ^k (−1)^k x^(k+1) −(x/((1+x)^2 )) =(2/(1+x))Σ_(i=0) ^∞ Σ_(k=0) ^∞ (−1)^(k+i) x^(k+i+1) −(x/((1+x)^2 )) =((2x)/((1+x)^3 ))−(x/((1+x)^2 )) =((x−x^2 )/((1+x)^3 )) −−−−−−−−−−−−−−−− Σ_(n=1) ^∞ n^2 (−1)^(n−1) x^n =Σ_(n=0) ^∞ (n^2 +2n+1)(−1)^n x^(n+1) =x((yD)^2 +2yD+1)∣_(y=−x) Σ_(n=0) ^∞ y^n =x(y^2 D^2 +3yD+1)∣_(y=−x) (1/(1−y)) =x[((2y^2 )/((1−y)^3 ))+((3y)/((1−y)^2 ))+(1/(1−y))]_(y=−x) =((x(1−x))/((1+x)^3 ))
$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{n}^{\mathrm{2}} \left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{x}^{\mathrm{n}} \\ $$$$=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)−\mathrm{n}\right)\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{x}^{\mathrm{n}} \\ $$$$=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{x}^{\mathrm{n}} −\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{n}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{x}^{\mathrm{n}} \\ $$$$=\mathrm{2}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{x}^{\mathrm{n}} −\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{x}^{\mathrm{n}} \\ $$$$=\mathrm{2}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{k}\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{k}−\mathrm{1}} \mathrm{x}^{\mathrm{n}+\mathrm{k}} −\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{k}−\mathrm{1}} \mathrm{x}^{\mathrm{n}+\mathrm{k}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{x}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{k}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{x}^{\mathrm{k}+\mathrm{1}} −\frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{x}}\underset{\mathrm{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}+\mathrm{i}} \mathrm{x}^{\mathrm{k}+\mathrm{i}+\mathrm{1}} −\frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2x}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }−\frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{x}−\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} } \\ $$$$−−−−−−−−−−−−−−−− \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{n}^{\mathrm{2}} \left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{x}^{\mathrm{n}} \\ $$$$=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{2n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{n}+\mathrm{1}} \\ $$$$=\mathrm{x}\left(\left(\mathrm{yD}\right)^{\mathrm{2}} +\mathrm{2yD}+\mathrm{1}\right)\mid_{\mathrm{y}=−\mathrm{x}} \underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{y}^{\mathrm{n}} \\ $$$$=\mathrm{x}\left(\mathrm{y}^{\mathrm{2}} \mathrm{D}^{\mathrm{2}} +\mathrm{3yD}+\mathrm{1}\right)\mid_{\mathrm{y}=−\mathrm{x}} \frac{\mathrm{1}}{\mathrm{1}−\mathrm{y}} \\ $$$$=\mathrm{x}\left[\frac{\mathrm{2y}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{3}} }+\frac{\mathrm{3y}}{\left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{y}}\right]_{\mathrm{y}=−\mathrm{x}} \\ $$$$=\frac{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} } \\ $$
Commented by mathdanisur last updated on 19/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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