Let (1+x)^{10} = sum_{r=0}^{10} c_{r} x^{r} a | vedclass

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(B) Given $P = \sum_{r=0}^{5} c_{2r} = c_{0} + c_{2} + c_{4} + c_{6} + c_{8} + c_{10}$.Since $c_{r} = {}^{10}C_{r}$, we have $P = {}^{10}C_{0} + {}^{1

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$8$

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(B) Given $P = \sum_{r=0}^{5} c_{2r} = c_{0} + c_{2} + c_{4} + c_{6} + c_{8} + c_{10}$.Since $c_{r} = {}^{10}C_{r}$, we have $P = {}^{10}C_{0} + {}^{10}C_{2} + {}^{10}C_{4} + {}^{10}C_{6} + {}^{10}C_{8} + {}^{10}C_{10} = 2^{10-1} = 2^{9}$.Given $Q = \sum_{r=0}^{3} d_{2r+1} = d_{1} + d_{3} + d_{5} + d_{7}$.Since $d_{r} = {}^{7}C_{r}$, we have $Q = {}^{7}C_{1} + {}^{7}C_{3} + {}^{7}C_{5} + {}^{7}C_{7} = 2^{7-1} = 2^{6}$.Therefore, $\frac{P}{Q} = \frac{2^{9}}{2^{6}} = 2^{9-6} = 2^{3} = 8$.

Let $(1+x)^{10} = \sum_{r=0}^{10} c_{r} x^{r}$ and $(1+x)^{7} = \sum_{r=0}^{7} d_{r} x^{r}$. If $P = \sum_{r=0}^{5} c_{2r}$ and $Q = \sum_{r=0}^{3} d_{2r+1}$, then $\frac{P}{Q}$ is equal to:

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