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Question Number 197880 by universe last updated on 02/Oct/23
find the sum of infinite series (1/2^1 )∙(1/3^2 ) + (1/2^2 )∙(1/3^4 )(1^2 +2^2 +3^2 ) + (1/2^3 )∙(1/3^6 )(1^2 +2^2 +3^2 +...+7^2 )+ (1/2^4 )∙(1/3^8 )(1^2 +2^2 +3^2 +...+15^2 )+........
$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{infinite}\:\mathrm{series} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{6}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+\mathrm{7}^{\mathrm{2}} \right)+ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{8}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+\mathrm{15}^{\mathrm{2}} \right)+…….. \\ $$
Answered by qaz last updated on 02/Oct/23
Σ_(i=1) ^∞ (1/(18^i ))(Σ_(j=1) ^(2^i −1) j^2 )=(1/6)Σ_(i=1) ^∞ (1/(18^i ))∙2^i (2^i −1)(2^(i+1) −1) =(1/6)Σ_(i=1) ^∞ (2∙((4/9))^i −3∙((2/9))^i +(1/9^i )) =(1/6)(2∙((4/9)/(1−(4/9)))−3∙((2/9)/(1−(2/9)))+((1/9)/(1−(1/9))))=((243)/(1680))
$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{18}^{{i}} }\left(\underset{{j}=\mathrm{1}} {\overset{\mathrm{2}^{{i}} −\mathrm{1}} {\sum}}{j}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{6}}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{18}^{{i}} }\centerdot\mathrm{2}^{{i}} \left(\mathrm{2}^{{i}} −\mathrm{1}\right)\left(\mathrm{2}^{{i}+\mathrm{1}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}\centerdot\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{{i}} −\mathrm{3}\centerdot\left(\frac{\mathrm{2}}{\mathrm{9}}\right)^{{i}} +\frac{\mathrm{1}}{\mathrm{9}^{{i}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{2}\centerdot\frac{\frac{\mathrm{4}}{\mathrm{9}}}{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{9}}}−\mathrm{3}\centerdot\frac{\frac{\mathrm{2}}{\mathrm{9}}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{9}}}+\frac{\frac{\mathrm{1}}{\mathrm{9}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}}\right)=\frac{\mathrm{243}}{\mathrm{1680}} \\ $$
Commented by universe last updated on 03/Oct/23
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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