Question

The infinite sum of $1+\frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+\frac{25}{7^4}+$ .......Is k. The Value of 27K is___

Answer 1

Let $S=1+\frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+\frac{25}{7^4}+$... multiplying both sides by the common ratio $\left(\frac{1}{7}\right)$, we get
$\frac{S}{7}=\frac{1}{7}+\frac{4}{7^2}+\frac{9}{7^4}+\frac{25}{7^5}+$... Subtracting the second equation from the first, we obtain
$\frac{6S}{7}=1+\frac{3}{7}+\frac{5}{7^2}+\frac{7}{7^3}+\frac{9}{7^4}+...$ again multiplying both sides by $\frac{1}{7}$ we get
$\frac{36S}{49}=1+\frac{2}{7}+\frac{2}{7^2}+\frac{2}{7^3}+\frac{2}{7^4}+....=1+\frac{2}{7}(1+\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...)\frac{36S}{49}=1+\frac{2}{7}\times \frac{7}{6}=\frac{4}{3}\Rightarrow S=\frac{49}{27}$