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

The objective is to determine the value of \( \alpha \) from the provided equation:

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

Underlying Principles:

This solution leverages the Binomial Theorem and its related properties. The essential equations are:

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

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

2. Sum of Binomial Coefficients:

\[ \sum_{k=0}^{n} {}^{n}C_k = 2^n \]

Consequently, \( \sum_{k=1}^{n} {}^{n}C_k = 2^n - 1 \).

Detailed Solution:

Step 1: Let the left-hand side (LHS) be represented by S. Simplify the term within the summation's parentheses.

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

Step 2: Substitute this simplified expression back into the summation and re-index. Setting \( k = r+1 \), the summation bounds for \( k \) become 1 to 11, and \( r = k-1 \).

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

Step 3: Decompose the summation into two distinct series.

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

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

\[ \sum_{k=1}^{11} {}^{11}C_k = 2^{11} - 1 \]

The first term simplifies to \( 10(2^{11} - 1) \).

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

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

The second term becomes \( 10 \left( \frac{11^{11}}{10^{11}} - 1 \right) = \frac{11^{11}}{10^{10}} - 10 \).

Step 6: Consolidate the results to determine 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 Calculation and Outcome:

Equate the computed LHS with the given RHS.

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

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

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

By comparing the numerators:

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

Therefore, \( \alpha = 20 \).