JEE Mains · Maths · STD 11 - 8. sequence and series
\( (\frac{1}{3}+\frac{4}{7})+(\frac{1}{3^{2}}+\frac{1}{3}\times\frac{4}{7}+\frac{4^{2}}{7^{2}})+(\frac{1}{3^{3}}+\frac{1}{3^{2}}\times\frac{4}{7}+\frac{1}{3}\times\frac{4^{2}}{7^{2}}+\frac{4^{3}}{7^{3}}) + \dots \) upto infinite terms is equal to -
- A \( \frac{5}{2} \)
- B \( \frac{7}{4} \)
- C \( \frac{4}{3} \)
- D \( \frac{6}{5} \)
Answer & Solution
Correct Answer
(A) \( \frac{5}{2} \)
Step-by-step Solution
Detailed explanation
Let \(a =\frac{4}{7}, b=\frac{1}{3}\) Multiply \(N ^{ r }\) and \(D ^{ r }\) by \(( a - b )=\frac{4}{7}-\frac{1}{3}=\frac{5}{21}\) \(\frac{1}{a-b}\left[\left(a^2-b^2\right)+\left(a^3-b^3\right)+\left(a^4-b^4\right)+\ldots \infty\right]\)…
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