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

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(B) આપેલ છે કે $P = \sum_{r=0}^{5} c_{2r} = c_{0} + c_{2} + c_{4} + c_{6} + c_{8} + c_{10}$.$c_{r} = {}^{10}C_{r}$ હોવાથી, $P = {}^{10}C_{0} + {}^{10}

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) આપેલ છે કે $P = \sum_{r=0}^{5} c_{2r} = c_{0} + c_{2} + c_{4} + c_{6} + c_{8} + c_{10}$.$c_{r} = {}^{10}C_{r}$ હોવાથી, $P = {}^{10}C_{0} + {}^{10}C_{2} + {}^{10}C_{4} + {}^{10}C_{6} + {}^{10}C_{8} + {}^{10}C_{10} = 2^{10-1} = 2^{9}$.આપેલ છે કે $Q = \sum_{r=0}^{3} d_{2r+1} = d_{1} + d_{3} + d_{5} + d_{7}$.$d_{r} = {}^{7}C_{r}$ હોવાથી, $Q = {}^{7}C_{1} + {}^{7}C_{3} + {}^{7}C_{5} + {}^{7}C_{7} = 2^{7-1} = 2^{6}$.તેથી, $\frac{P}{Q} = \frac{2^{9}}{2^{6}} = 2^{9-6} = 2^{3} = 8$.

ધારો કે $(1+x)^{10} = \sum_{r=0}^{10} c_{r} x^{r}$ અને $(1+x)^{7} = \sum_{r=0}^{7} d_{r} x^{r}$. જો $P = \sum_{r=0}^{5} c_{2r}$ અને $Q = \sum_{r=0}^{3} d_{2r+1}$ હોય, તો $\frac{P}{Q}$ ની કિંમત શોધો:

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