If $3 \times { }^5 C_0 + 8 \times { }^5 C_1 + 13 \times { }^5 C_2 + 18 \times { }^5 C_3 + 23 \times { }^5 C_4 + 28 \times { }^5 C_5 = k \times 2^n$,where $n$ is a power of $2$,find $k$. Specifically,if $3 \times { }^5 C_0 + 8 \times { }^5 C_1 + 13 \times { }^5 C_2 + 18 \times { }^5 C_3 + 23 \times { }^5 C_4 + 28 \times { }^5 C_5 = k \times 2^4$,then $k=$
- A$33$
- B$37$
- C$31$
- D$30$
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Similar Questions
$\sum_{k=0}^{20} \left({}^{20}C_{k}\right)^{2}$ is equal to :
DifficultJEE MAIN 2021
View SolutionIf $\sum_{r=0}^{20} {}^{20+r}C_r = \frac{p}{q} {}^{40}C_{20}$ and $GCD(p, q) = 1$,then $p^2 - q^2 =$
If $(1 + x)^{15} = C_0 + C_1x + C_2x^2 + ...... + C_{15}x^{15},$ then $C_2 + 2C_3 + 3C_4 + .... + 14C_{15} = $
DifficultIIT 1966
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:
$\binom{10}{1} + \binom{10}{2} + \binom{11}{3} + \binom{12}{4} + \binom{13}{5} = \dots$
Easy
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