JEE 2022 — Mathematics PYQ

Solution

1. Identify the pattern in the numerators:

Let the numerators be $T_n$: $1,5,12,22,35,51,70\dots$

• First differences: $4,7,10,13,16,19\dots$ (an AP with $d=3$)

• Second differences: $3,3,3,3\dots$ (constant)

  Since the second difference is constant, this is an Arithmetico-Geometric Series of the second order.

2. Summation Method:

Let $S=1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\dots$

Multiply by common ratio $\frac{1}{6}$:

$\frac{1}{6}S=\frac{1}{6}+\frac{5}{6^2}+\frac{12}{6^3}+\dots$

Subtracting the two:

3. Solve the inner AP-GP:

Let $S_1=\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+\dots$

$\frac{1}{6}{S}_1=\frac{4}{6^2}+\frac{7}{6^3}+\dots$

$\frac{5}{6}{S}_1=\frac{4}{6}+\frac{3}{6^2}+\frac{3}{6^3}+\cdots =\frac{4}{6}+\frac{3/36}{1-1/6}=\frac{4}{6}+\frac{3}{30}=\frac{23}{30}$

$S_1=\frac{23}{30}\times \frac{6}{5}=\frac{23}{25}$

4. Final Calculation:

Correct Option: (C)