Question

If \[ \sum_{r=0}^{10} \left( \frac{10^{r+1} - 1}{10^r} \right) \cdot {^{11}C_{r+1}} = \frac{\alpha^{11} - 11^{11}}{10^{10}}, \] then \( \alpha \) is equal to:

• A. 15
• B. 11
• C. 24
• D. 20

Hint:

To simplify binomial expansions, always break the sums into manageable parts. Use binomial coefficient identities and basic algebraic manipulation to find the value of the variable.

Answer 1

The Correct Option is D

We are asked to find the value of \( \alpha \) from the given equation:

\[ \sum_{r=0}^{10} \left( \frac{10^{r+1}-1}{10^r} \right) \cdot {}^{11}C_{r+1} = \frac{\alpha^{11} - 11^{11}}{10^{10}} \]

Concept Used:

The solution uses the Binomial Theorem and its consequences. The key formulas are:

1. Binomial expansion of \( (1+x)^n \):

\[ (1+x)^n = \sum_{k=0}^{n} {}^{n}C_k x^k = {}^{n}C_0 + {}^{n}C_1 x + \dots + {}^{n}C_n x^n \]

2. Sum of binomial coefficients:

\[ \sum_{k=0}^{n} {}^{n}C_k = {}^{n}C_0 + {}^{n}C_1 + \dots + {}^{n}C_n = 2^n \]

From this, it follows that \( \sum_{k=1}^{n} {}^{n}C_k = 2^n - {}^{n}C_0 = 2^n - 1 \).

Step-by-Step Solution:

Step 1: Let the left-hand side (LHS) be denoted by S. First, simplify the term inside the parentheses in the summation.

\[ \frac{10^{r+1}-1}{10^r} = \frac{10^{r+1}}{10^r} - \frac{1}{10^r} = 10 - \frac{1}{10^r} \]

Step 2: Substitute this simplified term back into the summation and perform a change of index. Let \( k = r+1 \). As \( r \) goes from 0 to 10, \( k \) goes from 1 to 11. Also, \( r = k-1 \).

\[ S = \sum_{r=0}^{10} \left( 10 - \frac{1}{10^r} \right) {}^{11}C_{r+1} = \sum_{k=1}^{11} \left( 10 - \frac{1}{10^{k-1}} \right) {}^{11}C_k \]

Step 3: Split the summation into two parts.

\[ S = \sum_{k=1}^{11} 10 \cdot {}^{11}C_k - \sum_{k=1}^{11} \frac{1}{10^{k-1}} \cdot {}^{11}C_k \] \[ S = 10 \sum_{k=1}^{11} {}^{11}C_k - \sum_{k=1}^{11} \frac{10}{10^k} \cdot {}^{11}C_k \] \[ S = 10 \sum_{k=1}^{11} {}^{11}C_k - 10 \sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k \]

Step 4: Evaluate the first summation using the sum of binomial coefficients identity.

\[ \sum_{k=1}^{11} {}^{11}C_k = {}^{11}C_1 + {}^{11}C_2 + \dots + {}^{11}C_{11} = (2^{11} - {}^{11}C_0) = 2^{11} - 1 \]

So, the first term is \( 10(2^{11} - 1) \).

Step 5: Evaluate the second summation using the binomial expansion of \( (1+x)^{11} \) with \( x = 1/10 \).

\[ \left(1 + \frac{1}{10}\right)^{11} = \sum_{k=0}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k = {}^{11}C_0 + \sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k \] \[ \left(\frac{11}{10}\right)^{11} = 1 + \sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k \] \[ \sum_{k=1}^{11} {}^{11}C_k \left(\frac{1}{10}\right)^k = \frac{11^{11}}{10^{11}} - 1 \]

So, the second term is \( 10 \left( \frac{11^{11}}{10^{11}} - 1 \right) = \frac{11^{11}}{10^{10}} - 10 \).

Step 6: Combine the results to find the value of S.

\[ S = 10(2^{11} - 1) - \left( \frac{11^{11}}{10^{10}} - 10 \right) \] \[ S = 10 \cdot 2^{11} - 10 - \frac{11^{11}}{10^{10}} + 10 = 10 \cdot 2^{11} - \frac{11^{11}}{10^{10}} \]

Final Computation & Result:

Now, we equate the calculated LHS with the given RHS.

\[ 10 \cdot 2^{11} - \frac{11^{11}}{10^{10}} = \frac{\alpha^{11} - 11^{11}}{10^{10}} \]

Rewrite the LHS with a common denominator of \( 10^{10} \).

\[ \frac{10^{10} \cdot (10 \cdot 2^{11}) - 11^{11}}{10^{10}} = \frac{\alpha^{11} - 11^{11}}{10^{10}} \] \[ \frac{10^{11} \cdot 2^{11} - 11^{11}}{10^{10}} = \frac{\alpha^{11} - 11^{11}}{10^{10}} \]

By comparing the numerators, we get:

\[ (10 \cdot 2)^{11} - 11^{11} = \alpha^{11} - 11^{11} \] \[ 20^{11} = \alpha^{11} \]

Therefore, the value of \( \alpha \) is 20.