Question Number 108967 by bobhans last updated on 20/Aug/20
$$\:\:\overset{\bigstar} {{b}}\overset{\bigstar} {{o}}\overset{\bigstar} {{b}}\overset{\sim} {{h}}\overset{\sim} {{a}}\overset{\sim} {{n}}\overset{\sim} {{s}} \\ $$$$\mathrm{1}+\frac{\mathrm{4}}{\mathrm{11}}+\frac{\mathrm{9}}{\mathrm{121}}+\frac{\mathrm{16}}{\mathrm{1331}}+\frac{\mathrm{25}}{\mathrm{14641}}+… \\ $$
Answered by john santu last updated on 20/Aug/20
$$\:\:\:\:\multimap\underset{\multimap} {{J}}\underset{\multimap} {{S}}\multimap \\ $$$${consider}\:{gemetric}\:{series}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:=\:\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \\ $$$$\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:=\:\frac{{d}}{{dx}}\left(\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \right) \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:=\:\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}{nx}^{{n}−\mathrm{1}} \\ $$$$\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:=\:\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}{nx}^{{n}} \\ $$$${diffrentiating}\:{both}\:{side}\: \\ $$$$\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:=\:\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} \\ $$$${letting}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{11}}\:{yields}\:{the}\:{desired}\:{result} \\ $$$$\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{2}} }{\mathrm{11}^{{n}−\mathrm{1}} }\:=\:\frac{\mathrm{363}}{\mathrm{250}} \\ $$
Commented by bobhans last updated on 20/Aug/20
$${thank}\:{you}\:{all}\:{master} \\ $$
Answered by Dwaipayan Shikari last updated on 20/Aug/20
$${S}=\mathrm{1}+\mathrm{4}{x}+\mathrm{9}{x}^{\mathrm{2}} +\mathrm{16}{x}^{\mathrm{3}} +….\:\:\:\left({x}=\frac{\mathrm{1}}{\mathrm{11}}\right) \\ $$$$−{xS}=\:−{x}−\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}{x}^{\mathrm{3}} −….. \\ $$$${S}\left(\mathrm{1}−{x}\right)=\mathrm{1}+\mathrm{3}{x}+\mathrm{5}{x}^{\mathrm{2}} +\mathrm{7}{x}^{\mathrm{3}} +…. \\ $$$${S}'=\mathrm{1}+\mathrm{3}{x}+\mathrm{5}{x}^{\mathrm{2}} +\mathrm{7}{x}^{\mathrm{3}} +….\:\:\:\:\left({S}'={S}\left(\mathrm{1}−{x}\right)\right. \\ $$$$−{xS}'=−{x}−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}{x}^{\mathrm{3}} −… \\ $$$${S}'\left(\mathrm{1}−{x}\right)=\mathrm{1}+\mathrm{2}\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…..\right) \\ $$$${S}'\left(\mathrm{1}−{x}\right)=\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{1}−{x}} \\ $$$${S}'=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${S}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\Rightarrow{S}=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\frac{\frac{\mathrm{12}}{\mathrm{11}}}{\mathrm{1000}}.\mathrm{1331}=\frac{\mathrm{363}}{\mathrm{250}} \\ $$$$ \\ $$
Answered by nimnim last updated on 20/Aug/20
$${Let}\:\:\:{S}=\mathrm{1}+\frac{\mathrm{4}}{\mathrm{11}}+\frac{\mathrm{9}}{\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{16}}{\mathrm{11}^{\mathrm{3}} }+\frac{\mathrm{25}}{\mathrm{11}^{\mathrm{4}} }+……..\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\mathrm{11}{S}=\mathrm{11}+\mathrm{4}+\frac{\mathrm{9}}{\mathrm{11}}+\frac{\mathrm{16}}{\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{25}}{\mathrm{11}^{\mathrm{3}} }+\frac{\mathrm{36}}{\mathrm{11}^{\mathrm{4}} }+…..\left(\mathrm{2}\right)\:\:\:\:\:\:\left\{\left(\mathrm{1}\right)×\mathrm{11}\right\} \\ $$$$\therefore\:\mathrm{10}{S}=\mathrm{14}+\frac{\mathrm{5}}{\mathrm{11}}+\frac{\mathrm{7}}{\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{9}}{\mathrm{11}^{\mathrm{3}} }+\frac{\mathrm{11}}{\mathrm{11}^{\mathrm{4}} }+…….\left(\mathrm{3}\right)\:\:\:\:\:\:\left\{\left(\mathrm{2}\right)−\left(\mathrm{1}\right)\right\} \\ $$$$\:{also},\:\frac{{S}}{\mathrm{11}}=\frac{\mathrm{1}}{\mathrm{11}}+\frac{\mathrm{4}}{\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{9}}{\mathrm{11}^{\mathrm{3}} }+\frac{\mathrm{16}}{\mathrm{11}^{\mathrm{4}} }+\frac{\mathrm{25}}{\mathrm{11}^{\mathrm{5}} }+….\left(\mathrm{4}\right)\:\:\:\:\:\left\{\left(\mathrm{1}\right)\boldsymbol{\div}\mathrm{11}\right\} \\ $$$$\therefore\:\frac{\mathrm{10}}{\mathrm{11}}{S}=\mathrm{1}+\frac{\mathrm{3}}{\mathrm{11}}+\frac{\mathrm{5}}{\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{7}}{\mathrm{11}^{\mathrm{3}} }+\frac{\mathrm{9}}{\mathrm{11}^{\mathrm{4}} }+……\left(\mathrm{5}\right)\:\:\:\:\:\left\{\left(\mathrm{1}\right)−\left(\mathrm{4}\right)\right\} \\ $$$$\frac{\mathrm{100}}{\mathrm{11}}{S}=\mathrm{13}+\frac{\mathrm{2}}{\mathrm{11}}+\frac{\mathrm{2}}{\mathrm{11}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{11}^{\mathrm{3}} }+………\:\:\left\{\left(\mathrm{3}\right)−\left(\mathrm{5}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{13}+\mathrm{2}\left[\frac{\frac{\mathrm{1}}{\mathrm{11}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{11}}}\right]=\mathrm{13}+\frac{\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{66}}{\mathrm{5}} \\ $$$$\therefore\:{S}=\frac{\mathrm{66}}{\mathrm{5}}×\frac{\mathrm{11}}{\mathrm{100}}=\frac{\mathrm{363}}{\mathrm{250}}\bigstar \\ $$