The value of -{ }^{15}C_{1} + 2 cdot { }^{15} | vedclass

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(B) Let $S = \sum_{r=1}^{15} (-1)^{r} r \cdot { }^{15}C_{r} + \sum_{k=1, k 	ext{ is odd}}^{11} { }^{14}C_{k}$.First, evaluate the summation $A = \sum

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$2^{13}-14$

Vedclass · Answer

(B) Let $S = \sum_{r=1}^{15} (-1)^{r} r \cdot { }^{15}C_{r} + \sum_{k=1, k 	ext{ is odd}}^{11} { }^{14}C_{k}$.First, evaluate the summation $A = \sum_{r=1}^{15} (-1)^{r} r \cdot { }^{15}C_{r}$.Using the identity $r \cdot { }^{n}C_{r} = n \cdot { }^{n-1}C_{r-1}$, we get $A = \sum_{r=1}^{15} (-1)^{r} 15 \cdot { }^{14}C_{r-1} = 15 \sum_{r=1}^{15} (-1)^{r} { }^{14}C_{r-1}$.Let $j = r-1$, then $A = 15 \sum_{j=0}^{14} (-1)^{j+1} { }^{14}C_{j} = -15 \sum_{j=0}^{14} (-1)^{j} { }^{14}C_{j}$.Since $\sum_{j=0}^{n} (-1)^{j} { }^{n}C_{j} = 0$ for $n \ge 1$, we have $A = -15(0) = 0$.Next, evaluate $B = { }^{14}C_{1} + { }^{14}C_{3} + \ldots + { }^{14}C_{11}$.We know that $\sum_{k 	ext{ is odd}} { }^{n}C_{k} = 2^{n-1}$.For $n=14$, $\sum_{k 	ext{ is odd}} { }^{14}C_{k} = { }^{14}C_{1} + { }^{14}C_{3} + \ldots + { }^{14}C_{13} = 2^{14-1} = 2^{13}$.Thus, $B = 2^{13} - { }^{14}C_{13} = 2^{13} - 14$.Therefore, the total sum is $A + B = 0 + 2^{13} - 14 = 2^{13} - 14$.

The value of $-{ }^{15}C_{1} + 2 \cdot { }^{15}C_{2} - 3 \cdot { }^{15}C_{3} + \ldots - 15 \cdot { }^{15}C_{15} + { }^{14}C_{1} + { }^{14}C_{3} + { }^{14}C_{5} + \ldots + { }^{14}C_{11}$ is

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