<p>The sum of the series \(2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \cdots + 62 \cdot {}^{20}C_{20}\) is equal to:</p>
Step-by-Step Solution
Key Concept: Recognize that coefficients 2, 5, 8, 11, ..., 62 form an arithmetic sequence (3k+2), allowing you to split the sum into a linear combination of binomial coefficient sums, which can be evaluated using (1+x)^n identities and differentiation.
<p><strong>Step 1:</strong> Identify the coefficient pattern. The coefficients are 2, 5, 8, 11, ..., 62, forming the sequence 3k+2 where k = 0, 1, 2, ..., 20.</p><p><strong>Step 2:</strong> Split the sum: $$\sum_{k=0}^{20} (3k+2)\binom{20}{k} = 3\sum_{k=0}^{20} k\binom{20}{k} + 2\sum_{k=0}^{20}\binom{20}{k}$$</p><p><strong>Step 3:</strong> Evaluate the second sum: $\sum_{k=0}^{20}\binom{20}{k} = 2^{20}$</p><p><strong>Step 4:</strong> For the first sum, use the identity $k\binom{20}{k} = 20\binom{19}{k-1}$:<br/>$$\sum_{k=0}^{20} k\binom{20}{k} = 20\sum_{k=1}^{20}\binom{19}{k-1} = 20 \cdot 2^{19}$$</p><p><strong>Step 5:</strong> Combine results: $$3 \cdot 20 \cdot 2^{19} + 2 \cdot 2^{20} = 60 \cdot 2^{19} + 4 \cdot 2^{19} = 64 \cdot 2^{19} = 2^6 \cdot 2^{19} = 2^{25}$$</p><p>∴ Answer: B</p>
Correct Answer: B