Question

The value of $^{27}C_1 - ^{13}C_1) + (^{27}C_2 - ^{13}C_2) + (^{27}C_3 - ^{13}C_3) + ... + (^{27}C_{13} - ^{13}C_{13})$ is

• A. $2^{26} - 2^{12}$
• B. $2^{27} - 2^{13}$
• C. $2^{13}(2^{13} - 1)$
• D. $2^{14}(2^{13} - 1)$

Answer 1

Answer: (C)

$2^{26} - 2^{13}$

Answer 2

The sum can be written as \sum_{k=1}^{13} (^{27}C_k - ^{13}C_k) = \sum_{k=1}^{13} ^{27}C_k - \sum_{k=1}^{13} ^{13}C_k. We know that \sum_{k=0}^{n} ^nC_k = 2^n. Also, \sum_{k=0}^{27} ^{27}C_k = 2^{27} and \sum_{k=0}^{13} ^{13}C_k = 2^{13}. Due to symmetry, $^{n}C_k = ^{n}C_{n-k}$. For $^{27}C_k$, the sum of terms from $k=1$ to $k=13$ is equal to the sum of terms from $k=14$ to $k=26$. So, 2^{27} = ^{27}C_0 + \sum_{k=1}^{13} ^{27}C_k + \sum_{k=14}^{26} ^{27}C_k + ^{27}C_{27} = 1 + \sum_{k=1}^{13} ^{27}C_k + \sum_{k=1}^{13} ^{27}C_k + 1. 2^{27} = 2 + 2 \sum_{k=1}^{13} ^{27}C_k \implies \sum_{k=1}^{13} ^{27}C_k = \frac{2^{27}-2}{2} = 2^{26}-1. For the second sum, 2^{13} = ^{13}C_0 + \sum_{k=1}^{13} ^{13}C_k = 1 + \sum_{k=1}^{13} ^{13}C_k. So, \sum_{k=1}^{13} ^{13}C_k = 2^{13}-1. The total sum is $(2^{26}-1) - (2^{13}-1) = 2^{26}-2^{13}$. This can be factored as $2^{13}(2^{13}-1)$.