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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 $P=\sum_{r=0}^{5} c_{2}$ and $Q=\sum_{r=0}^{3} d_{2 r+1},$ then $\frac{P}{Q}$ is equal
to
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If $P=\sum_{r=0}^{5} c_{2}$ and $Q=\sum_{r=0}^{3} d_{2 r+1},$ then $\frac{P}{Q}$ is equal
to
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Verified Answer
The correct answer is:
8
$P=\sum_{r=0}^{5} C_{2 r}$
$={ }^{10} C_{0}+{ }^{10} C_{2}+\quad+{ }^{10} C_{10}=\frac{2^{0}}{2}=2^{9}$
$Q=\sum_{r=0}^{3} d_{2 r+1}=d_{1}+d_{3}+d_{5}+d_{7}$
$={ }^{7} C_{1}+{ }^{7} C_{3}+{ }^{7} C_{5}+{ }^{7} C_{7}=\frac{2^{7}}{2}=2^{6}$
$\therefore \frac{P}{\Omega}=\frac{2^{9}}{3^{6}}=2^{3}=8$
$={ }^{10} C_{0}+{ }^{10} C_{2}+\quad+{ }^{10} C_{10}=\frac{2^{0}}{2}=2^{9}$
$Q=\sum_{r=0}^{3} d_{2 r+1}=d_{1}+d_{3}+d_{5}+d_{7}$
$={ }^{7} C_{1}+{ }^{7} C_{3}+{ }^{7} C_{5}+{ }^{7} C_{7}=\frac{2^{7}}{2}=2^{6}$
$\therefore \frac{P}{\Omega}=\frac{2^{9}}{3^{6}}=2^{3}=8$
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