Question

If $ \sum_{r=0}^{10} \left( 10^{r+1} - 1 \right)$ $\,$\(\binom{10}{r} = \alpha^{11} - 1 \), 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 begin with the given sum: \[ \sum_{r=0}^{10} \left( 10^{r+1} - 1 \right) \binom{10}{r} \] Now, expanding the terms: \[ = \sum_{r=0}^{10} \left( 10^{r+1} - 10 \right) \binom{10}{r} \] This gives us two separate sums: \[ \sum_{r=0}^{10} \left( 10^{r+1} \right) \binom{10}{r} - \sum_{r=0}^{10} 10 \binom{10}{r} \] Now, evaluating both parts separately: \[ \sum_{r=0}^{10} 10^{r+1} \binom{10}{r} = 10 \sum_{r=0}^{10} 10^r \binom{10}{r} \] \[ \sum_{r=0}^{10} 10 \binom{10}{r} = 10 \left( \sum_{r=0}^{10} \binom{10}{r} \right) \] Using the binomial expansion for the sum of binomial coefficients: \[ \sum_{r=0}^{10} \binom{10}{r} = 2^{10} = 1024 \] Now, continuing: \[ 10 \sum_{r=0}^{10} 10^r \binom{10}{r} = 10^{11} - 1 \] Finally, simplifying: \[ 10^{11} - 1 = \alpha^{11} - 1 \] Hence, \( \alpha = 20 \).