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 $\frac{P}{2Q}$ is equal to
- A$2$
- B$4$
- C$8$
- D$16$
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The mean of the values $0, 1, 2, \dots, n$ having corresponding weights $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$ respectively is
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View SolutionIf $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n$,then $C_0+2 C_1+3 C_2+\ldots+(n+1) C_n$ is equal to
DifficultTS EAMCET 2001
View Solution$^{10}C_1 + ^{10}C_3 + ^{10}C_5 + ^{10}C_7 + ^{10}C_9 = $
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View SolutionMatch the expressions in List-$I$ with their values in List-$II$ for the expansion $(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.
List-$I$ List-$II$ $(A)$ $a_0 + a_2 + \ldots + a_{2n}$ $(I)$ $n \cdot 3^{n-1}$ $(B)$ $a_1 + a_3 + \ldots + a_{2n-1}$ $(II)$ $n \cdot 3^n$ $(C)$ $a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$ $(III)$ $\frac{1}{2}(3^n + 1)$ $(IV)$ $\frac{1}{2}(3^n - 1)$
The correct match is:
| List-$I$ | List-$II$ |
| $(A)$ $a_0 + a_2 + \ldots + a_{2n}$ | $(I)$ $n \cdot 3^{n-1}$ |
| $(B)$ $a_1 + a_3 + \ldots + a_{2n-1}$ | $(II)$ $n \cdot 3^n$ |
| $(C)$ $a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$ | $(III)$ $\frac{1}{2}(3^n + 1)$ |
| $(IV)$ $\frac{1}{2}(3^n - 1)$ |
The correct match is:
If $^nC_0 + \frac{1}{2} ^nC_1 + \frac{1}{3} ^nC_2 + \dots + \frac{1}{n+1} ^nC_n = \frac{1023}{10}$, then $n =$ ?
DifficultMHT CET 2026
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