Question

In the usual notation, \(\frac{^nC_1}{2} + \frac{^nC_2}{3} + \cdots + \frac{^nC_n}{n+1}\) is equal to

• A. \(\frac{2^{n+1}-1}{n+1}\)
• B. \(\frac{2^{n+1}-n-1}{n+1}\)
• C. \(\frac{2^{n+1}-n}{n+1}\)
• D. None of these

Hint:

Always check starting index (r=0 or r=1) in summation problems.

Answer 1

The Correct Option is D

Concept: \[ \int_0^1 (1+x)^n dx \]

Step 1: Expand binomial.
\[ (1+x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r \]

Step 2: Integrate.
\[ \int_0^1 (1+x)^n dx = \sum_{r=0}^{n} \frac{\binom{n}{r}}{r+1} \] \[ = \frac{2^{n+1}-1}{n+1} \]

Step 3: Adjust required sum.
Given sum starts from \(r=1\), so subtract \(r=0\) term: \[ \frac{\binom{n}{0}}{1} = 1 \] \[ \Rightarrow \sum_{r=1}^{n} \frac{\binom{n}{r}}{r+1} = \frac{2^{n+1}-1}{n+1} - 1 \] \[ = \frac{2^{n+1}-1-(n+1)}{n+1} = \frac{2^{n+1}-n-2}{n+1} \]

Step 4: Conclusion.
This expression is not among the given options.