WBJEE · Maths · Binomial Theorem
\(\begin{array}{llll}\text { 3. The sum } & \text { of } & \text { the } & \text { series }\end{array}\) \(1+\frac{1}{2}{ }^{n} C_{1}+\frac{1}{3}{ }^{n} C_{2}+\quad+\frac{1}{n+1}{ }^{n} C_{n}\) is equal to
- A \(\frac{2^{n+1}-1}{n+1}\)
- B \(\frac{3\left(2^{n}-1\right)}{2 n}\)
- C \(\frac{2^{n}+1}{n+1}\)
- D \(\frac{2^{n}+1}{2 n}\)
Answer & Solution
Correct Answer
(A) \(\frac{2^{n+1}-1}{n+1}\)
Step-by-step Solution
Detailed explanation
\(1+\frac{1}{2}{ }^{n} C_{1}+\frac{1}{3}{ }^{n} C_{2}+\quad+\frac{1}{n+1}{ }^{n} C_{n}\) \(=\frac{1}{n+1}\left[(n+1)+\frac{(n+1) n}{21}\right.\) \(+\frac{(n+1) n(n-1)}{3 !}+(1]\) \(=\frac{1}{n+1}\left[^{n+1} C_{1}+{ }^{n+1} C_{2}+\quad+{ }^{n+1} C_{n+1}\right]\)…
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