JEE Main 2023 — Sequences And Series Question with Solution

We can rewrite the given series as follows :

$$S=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{9}\right)+\left(\frac{1}{8}-\frac{1}{12}+\frac{1}{18}-\frac{1}{27}\right)+\left(\frac{1}{16}-\frac{1}{24}+\frac{1}{36}-\frac{1}{54}+\frac{1}{81}\right)+\dots$$

The first few terms of the series are :

$$S=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots \right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\dots \right)+\left(\frac{1}{9}+\frac{1}{18}+\frac{1}{36}+\dots \right)-\left(\frac{1}{27}+\frac{1}{54}+\dots \right)+\dots$$

We can now see that each group of terms forms a geometric series with a common ratio of $\frac{1}{2}$:

$$S=\frac{\frac{1}{2}}{1-\frac{1}{2}}-\frac{\frac{1}{3}}{1-\frac{1}{2}}+\frac{\frac{1}{9}}{1-\frac{1}{2}}-\frac{\frac{1}{27}}{1-\frac{1}{2}}+\dots$$

The series can be rewritten as :

$$S=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\dots \right)$$

Now, we can simplify and rewrite the series inside the parentheses as:

$$S=2\left[\frac{1}{2}+(-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\dots \right)]$$

The series inside the parentheses is an infinite geometric series with the first term $a=-\frac{1}{3}$ and the common ratio $r=-\frac{1}{3}$:

$$S=2\left(\frac{1}{2}+\frac{a}{1-r}\right)=2\left(\frac{1}{2}+\frac{-\frac{1}{3}}{1+\frac{1}{3}}\right)=2\left(\frac{1}{2}+\frac{-\frac{1}{3}}{\frac{4}{3}}\right)$$

= $$2\left(\frac{1}{2}-\frac{1}{3}\times \frac{3}{4}\right)$$

$$=2\left(\frac{1}{2}-\frac{1}{4}\right)=2\left(\frac{1}{4}\right)=\frac{1}{2}$$

Thus, the sum of the series is $\frac{1}{2}$, and $\alpha =1$ and $\beta =2$ are co-prime.

Therefore, $\alpha +3\beta =1+6=7$