JEE Main 2026 Jan 24 Shift 2, Mathematics Q7: Summation of Series

Let $a=\frac{4}{7}$, $b=\frac{1}{3}$. The $n$-th group is $\sum _{k=0}^n{a}^k{b}^{n-k}=\frac{a^{n+1}-b^{n+1}}{a-b}$ (geometric in two terms).

Summing over $n\ge 1$:

$S=\frac{1}{a-b}\left[{\sum }_{n=1}^{\infty }{a}^{n+1}-{\sum }_{n=1}^{\infty }{b}^{n+1}\right]=\frac{1}{a-b}\left[\frac{a^2}{1-a}-\frac{b^2}{1-b}\right]$.

Compute: $a-b=\frac{4}{7}-\frac{1}{3}=\frac{5}{21}$; $\frac{a^2}{1-a}=\frac{16/49}{3/7}=\frac{16}{21}$; $\frac{b^2}{1-b}=\frac{1/9}{2/3}=\frac{1}{6}$.

$\frac{a^2}{1-a}-\frac{b^2}{1-b}=\frac{16}{21}-\frac{1}{6}=\frac{96-21}{126}=\frac{75}{126}=\frac{25}{42}$.

$S=\frac{21}{5}\cdot \frac{25}{42}=\frac{525}{210}=\frac{5}{2}$.