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

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(B) दिया गया है $P = \sum_{r=0}^{5} c_{2r} = c_{0} + c_{2} + c_{4} + c_{6} + c_{8} + c_{10}$.चूँकि $c_{r} = {}^{10}C_{r}$, इसलिए $P = {}^{10}C_{0} + {

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) दिया गया है $P = \sum_{r=0}^{5} c_{2r} = c_{0} + c_{2} + c_{4} + c_{6} + c_{8} + c_{10}$.चूँकि $c_{r} = {}^{10}C_{r}$, इसलिए $P = {}^{10}C_{0} + {}^{10}C_{2} + {}^{10}C_{4} + {}^{10}C_{6} + {}^{10}C_{8} + {}^{10}C_{10} = 2^{10-1} = 2^{9}$.दिया गया है $Q = \sum_{r=0}^{3} d_{2r+1} = d_{1} + d_{3} + d_{5} + d_{7}$.चूँकि $d_{r} = {}^{7}C_{r}$, इसलिए $Q = {}^{7}C_{1} + {}^{7}C_{3} + {}^{7}C_{5} + {}^{7}C_{7} = 2^{7-1} = 2^{6}$.अतः, $\frac{P}{Q} = \frac{2^{9}}{2^{6}} = 2^{9-6} = 2^{3} = 8$.

मान लीजिए $(1+x)^{10} = \sum_{r=0}^{10} c_{r} x^{r}$ और $(1+x)^{7} = \sum_{r=0}^{7} d_{r} x^{r}$ है। यदि $P = \sum_{r=0}^{5} c_{2r}$ और $Q = \sum_{r=0}^{3} d_{2r+1}$ है, तो $\frac{P}{Q}$ का मान ज्ञात कीजिए:

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