The value of $\log_8\!\left(2^{50}\binom{50}{0} + 2^{49}\binom{51}{1} + 2^{48}\binom{52}{2} + \cdots + 2^0\binom{100}{50}\right)$ is equal to:
Step-by-Step Solution
Key Concept: The sum $\displaystyle\sum_{k=0}^{50}2^{50-k}\binom{50+k}{k} = 2^{50}\sum_{k=0}^{50}\frac{1}{2^k}\binom{50+k}{k}$. Use the identity: $\sum_{k=0}^{n}\binom{n+k}{k}x^k = \frac{1}{(1-x)^{n+1}}$ evaluated appropriately, or recognize as a power of $2$.
Sum $=2^{100}$ (by combinatorial identity). $\log_8(2^{100})=\frac{100\log 2}{3\log 2}=\frac{100}{3}=\mathbf{33.33}$.
Correct Answer: 33.33