Question Number 203357 by Calculusboy last updated on 17/Jan/24
$$\boldsymbol{{For}}\:\boldsymbol{{the}}\:\boldsymbol{{series}}\:\mathrm{5}−\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{4}}−\frac{\mathrm{5}}{\mathrm{8}}+\centerdot\centerdot\centerdot+\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}−\mathrm{1}} \mathrm{5}}{\mathrm{2}^{\boldsymbol{{n}}−\mathrm{1}} } \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{an}}\:\boldsymbol{{expression}}\:\boldsymbol{{for}}\:\boldsymbol{{the}}\:\boldsymbol{{sum}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{first}} \\ $$$$\boldsymbol{{n}}\:\boldsymbol{{terms}}.\:\boldsymbol{{Also}}\:\boldsymbol{{if}}\:\boldsymbol{{the}}\:\boldsymbol{{series}}\:\boldsymbol{{converges}}, \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{sum}}\:\boldsymbol{{to}}\:\infty. \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 17/Jan/24
![G.P. with q=−(1/2) S_n =((5(1−(−(1/2))^n ))/(1−(−(1/2))))=((10)/3)[1−(((−1)^n )/2^n )] lim_(n→∞) S_n =((10)/3)](https://www.tinkutara.com/question/Q203363.png)
$${G}.{P}.\:{with}\:{q}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${S}_{{n}} =\frac{\mathrm{5}\left(\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \right)}{\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{10}}{\mathrm{3}}\left[\mathrm{1}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\right] \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} =\frac{\mathrm{10}}{\mathrm{3}} \\ $$
Commented by Calculusboy last updated on 17/Jan/24
$$\boldsymbol{{nice}}\:\boldsymbol{{solution}}\:\boldsymbol{{sir}} \\ $$