Q32 / 100 · BITSAT 2024 (Memory Based Paper 2) · Mathematics
The sum of the infinite series
$1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\frac{35}{6^4}+\frac{51}{6^5}+\frac{70}{6^6}+\ldots .$. is equal
to:
- A$\frac{425}{216}$
- B$\frac{429}{216}$
- Correct answerC$\frac{288}{125}$
- D$\frac{280}{125}$
Solution
Let $S=1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\frac{35}{6^4}+\ldots$ .....(i)
$\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+\frac{12}{6^3}+\frac{22}{6^4}+\ldots$ ....(ii)
On subtracting (i) from (ii)
$\frac{5}{6} S=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+\frac{13}{6^4}+\cdots$
$\frac{5}{36} S=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+\frac{10}{6^4}+\frac{13}{6^5}+\cdots$
on subtraction
$\frac{25}{36} S=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+\ldots=\frac{8}{5}$
$S=\frac{8}{5} \times \frac{36}{25}=\frac{288}{125}$
Hence, the correct answer is option: 'C'.