Sequences & Series
Infinite Series of GP Bracket Sums
nta_pyq_2026_jan
Grade None
Question:
$\left(\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3^2}+\dfrac{1}{3}\times\dfrac{4}{7}+\dfrac{4^2}{7^2}\right)+\left(\dfrac{1}{3^3}+\dfrac{1}{3^2}\times\dfrac{4}{7}+\dfrac{1}{3}\times\dfrac{4^2}{7^2}+\dfrac{4^3}{7^3}\right)+\cdots$ upto infinite terms, is equal to
\dfrac{4}{3}
\dfrac{5}{6}
\dfrac{5}{2}
\dfrac{7}{4}
Step-by-Step Solution
Key Concept: The $n$th bracket $=\sum_{k=0}^{n}(1/3)^{n-k}(4/7)^k=\tfrac{(4/7)^{n+1}-(1/3)^{n+1}}{4/7-1/3}=\tfrac{21}{5}\left[(4/7)^{n+1}-(1/3)^{n+1}\right]$.
$\dfrac{5}{2}$.
Correct Answer: 3