If $7=5+\frac17(5+\alpha)+\frac1{7^2}(5+2\alpha)+\frac1{7^3}(5+3\alpha)+\dots\infty$, then the value of $\alpha$ is :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\sum_{k\ge1}\frac5{7^k}=\frac56$ and $\sum_{k\ge1}\frac{k\alpha}{7^k}=\alpha\cdot\frac{1/7}{(6/7)^2}=\frac{7\alpha}{36}$. So $7=5+\frac56+\frac{7\alpha}{36}\Rightarrow\frac76=\frac{7\alpha}{36}\Rightarrow\alpha=6$. Answer (2).

Q. If $7=5+\frac17(5+\alpha)+\frac1{7^2}(5+2\alpha)+\frac1{7^3}(5+3\alpha)+\dots\infty$, then the value of $\alpha$ is :

Explanation

Reviewed by Dr. Abubacker

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