Let $c_0, c_1, c_2, \ldots, c_n$ be the binomial coefficients in the expansion of $(1+x)^n$. If $S_{n+1} = 5 \cdot c_0 + 8 \cdot c_1 + 11 \cdot c_2 + \ldots$ ($n+1$ terms),then $S_{11} =$
- A$18944$
- B$17920$
- C$20480$
- D$40960$
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Similar Questions
Match 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 $n$ is an integer greater than $1$,then $a - ^nC_1(a - 1) + ^nC_2(a - 2) + \dots + (-1)^n(a - n) = $
MediumIIT 1972
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 SolutionIf $C_r = ^{100}C_r$,then the value of $1 \cdot C_0^2 - 2 \cdot C_1^2 + 3 \cdot C_2^2 - 4 \cdot C_3^2 + \dots + 101 \cdot C_{100}^2$ is equal to:
If $1^2 \cdot ^{20}C_1 + 2^2 \cdot ^{20}C_2 + 3^2 \cdot ^{20}C_3 + \dots + 20^2 \cdot ^{20}C_{20} = A(2^\beta)$,then the ordered pair $(A, \beta)$ is equal to
DifficultJEE MAIN 2019
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