Question:
The value of $n$ for which the sum
\[
\frac{{}^nC_0}{2^n}
+
2\frac{{}^nC_1}{2^n}
+
3\frac{{}^nC_2}{2^n}
+\cdots+
(n+1)\frac{{}^nC_n}{2^n}
=16
\]
is
The value of $n$ for which the sum
\[
\frac{{}^nC_0}{2^n}
+
2\frac{{}^nC_1}{2^n}
+
3\frac{{}^nC_2}{2^n}
+\cdots+
(n+1)\frac{{}^nC_n}{2^n}
=16
\]
is
Show Hint
Memorize: $\sum r\,{}^nC_r=n2^{n-1}$ and $\sum {}^nC_r=2^n$.
Updated On: Jun 3, 2026
- $10$
- $20$
- $25$
- $30$
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Concept
Use standard binomial identities involving weighted binomial coefficients.
Step 2: Meaning
Let \[ S=\sum_{r=0}^{n}(r+1)\frac{{}^nC_r}{2^n}. \] Then \[ S=\frac{1}{2^n} \left( \sum_{r=0}^{n} r\,{}^nC_r + \sum_{r=0}^{n} {}^nC_r \right). \]
Step 3: Analysis
Using identities, \[ \sum_{r=0}^{n} {}^nC_r=2^n, \] and \[ \sum_{r=0}^{n} r\,{}^nC_r=n2^{\,n-1}. \] Therefore, \[ S=\frac{n2^{n-1}+2^n}{2^n} =\frac{n}{2}+1. \] Given \[ \frac{n}{2}+1=16. \] Hence, \[ \frac{n}{2}=15 \] and \[ n=30. \]
Step 4: Conclusion
Therefore the required value is $30$.
Final Answer: (D)
Use standard binomial identities involving weighted binomial coefficients.
Step 2: Meaning
Let \[ S=\sum_{r=0}^{n}(r+1)\frac{{}^nC_r}{2^n}. \] Then \[ S=\frac{1}{2^n} \left( \sum_{r=0}^{n} r\,{}^nC_r + \sum_{r=0}^{n} {}^nC_r \right). \]
Step 3: Analysis
Using identities, \[ \sum_{r=0}^{n} {}^nC_r=2^n, \] and \[ \sum_{r=0}^{n} r\,{}^nC_r=n2^{\,n-1}. \] Therefore, \[ S=\frac{n2^{n-1}+2^n}{2^n} =\frac{n}{2}+1. \] Given \[ \frac{n}{2}+1=16. \] Hence, \[ \frac{n}{2}=15 \] and \[ n=30. \]
Step 4: Conclusion
Therefore the required value is $30$.
Final Answer: (D)
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