Question

If \(C_r={}^{10}C_r\), then \[ 2C_0+\frac{2^2}{2}C_1+\frac{2^3}{3}C_2+\cdots+\frac{2^{11}}{11}C_{10} \] is equal to:

• A. \(\dfrac{3^{11}-1}{11}\)
• B. \(\dfrac{2^{11}-1}{11}\)
• C. \(\dfrac{3^{11}-1}{12}\)
• D. \(\dfrac{2^{11}-1}{12}\)

Hint:

Use the identity \[ \frac{{}^{n}C_r}{r+1}=\frac{{}^{n+1}C_{r+1}}{n+1} \] when a binomial coefficient is divided by \(r+1\).

Answer 1

The Correct Option is A

Concept:
This question is based on binomial coefficients and binomial expansion. Given: \[ C_r={}^{10}C_r \] We need to evaluate: \[ 2C_0+\frac{2^2}{2}C_1+\frac{2^3}{3}C_2+\cdots+\frac{2^{11}}{11}C_{10} \] The general term of the series is: \[ \frac{2^{r+1}}{r+1}C_r \] Since: \[ C_r={}^{10}C_r \] the general term becomes: \[ \frac{2^{r+1}}{r+1}{}^{10}C_r \]

Step 1: Write the series in summation form.
\[ S=\sum_{r=0}^{10}\frac{2^{r+1}}{r+1}{}^{10}C_r \]

Step 2: Use the binomial coefficient identity.
We use: \[ \frac{{}^{10}C_r}{r+1}=\frac{{}^{11}C_{r+1}}{11} \] This identity is useful because it converts the denominator \(r+1\) into a binomial coefficient of order \(11\). So: \[ \frac{2^{r+1}}{r+1}{}^{10}C_r = \frac{2^{r+1}}{11}{}^{11}C_{r+1} \]

Step 3: Substitute into the series.
\[ S=\sum_{r=0}^{10}\frac{2^{r+1}}{11}{}^{11}C_{r+1} \] \[ S=\frac{1}{11}\sum_{r=0}^{10}2^{r+1}{}^{11}C_{r+1} \]

Step 4: Change the index.
Let: \[ k=r+1 \] When: \[ r=0,\quad k=1 \] When: \[ r=10,\quad k=11 \] Therefore: \[ S=\frac{1}{11}\sum_{k=1}^{11}2^k{}^{11}C_k \]

Step 5: Use binomial expansion.
By binomial theorem: \[ (1+2)^{11}=\sum_{k=0}^{11}{}^{11}C_k2^k \] So: \[ 3^{11}={}^{11}C_0 2^0+\sum_{k=1}^{11}{}^{11}C_k2^k \] Since: \[ {}^{11}C_0 2^0=1 \] we get: \[ \sum_{k=1}^{11}{}^{11}C_k2^k=3^{11}-1 \]

Step 6: Find the final value.
\[ S=\frac{1}{11}(3^{11}-1) \] \[ S=\frac{3^{11}-1}{11} \] Hence, the correct answer is: \[ \boxed{(A)\ \frac{3^{11}-1}{11}} \]