$\left(\frac{1}{3}+\frac{4}{7}\right)+\left(\frac{1}{{3}^{2}}+\frac{1}{3}\times \frac{4}{7}+\frac{{4}^{2}}{{7}^{2}}\right)+$$\left(\frac{1}{{3}^{3}}+\frac{1}{{3}^{2}}\times \frac{4}{7}+\frac{1}{3}\times \frac{{4}^{2}}{{7}^{2}}+\frac{{4}^{3}}{{7}^{3}}\right)+...$upto infinite terms, is equal to

Question

YESPYQ · Accepted Answer

Correct answer: d) $\frac{5}{2}$. Let $a=\frac{4}{7},b=\frac{1}{3}$ Therefore we get $S=\left(a+b\right)+\left({a}^{2}+ab+{b}^{2}\right)+\left({a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3}\right)+...\infty$ Multiply numerator and denominator by $(a – b)$ $=\frac{4}{7}−\frac{1}{3}=\frac{5}{21}$ $S=\frac{1}{a−b}\left[\left({a}^{2}−{b}^{2}\right)+\left({a}^{3}−{b}^{3}\right)+\left({a}^{4}−{b}^{4}\right)+\ldots .∞\right]$ $S=\frac{1}{a−b}\left[\frac{{a}^{2}}{1−a}−\frac{{b}^{2}}{1−b}\right]$ $=\frac{21}{5}\left[\frac{\frac{16}{49}}{1−\frac{4}{7}}−\frac{\frac{1}{9}}{1−\frac{1}{3}}\right]$ $=\frac{21}{5}\left[\frac{16}{21}−\frac{1}{6}\right]=\frac{21}{5}\left[\frac{96−21}{21\cdot 6}\right]$ $=\frac{75}{5\cdot 6}=\frac{15}{6}=\frac{5}{2}$

\(\left(\frac{1}{3}+\frac{4}{7}\right)+\left(\frac{1}{{3}^{2}}+\frac{1}{3}\times \frac{4}{7}+\frac{{4}^{2}}{{7}^{2}}\rig…

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