TS EAMCET · Maths · Binomial Theorem
\(\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k}\left({ }^k C_r\right)\) is equal to
- A \(\frac{1}{3}\)
- B \(\frac{2}{3}\)
- C \(1\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k}\left({ }^k C_r\right) \\ = & \sum_{k=1}^{\infty} \frac{1}{3^k}\left({ }^k C_0+{ }^k C_1+{ }^k C_2+\ldots+{ }^k C_k\right) \\ = & \sum_{k=1}^{\infty} \frac{2^k}{3^k}=\sum_{k=1}^{\infty}\left(\frac{2}{3}\right)^k \\…
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