(C) The given sum is $S_{n+1} = \sum_{k=0}^{n} (3k+5) C_k$,where $C_k = \binom{n}{k}$.For $n=10$,we have $S_{11} = \sum_{k=0}^{10} (3k+5) C_k$.We know

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(C) The given sum is $S_{n+1} = \sum_{k=0}^{n} (3k+5) C_k$,where $C_k = \binom{n}{k}$.For $n=10$,we have $S_{11} = \sum_{k=0}^{10} (3k+5) C_k$.We know that $\sum_{k=0}^{n} C_k = 2^n$ and $\sum_{k=0}^{n} k C_k = n 2^{n-1}$.Thus,$S_{11} = 3 \sum_{k=0}^{10} k C_k + 5 \sum_{k=0}^{10} C_k$.Substituting $n=10$: $S_{11} = 3(10 \cdot 2^9) + 5(2^{10})$.$S_{11} = 30 \cdot 512 + 5 \cdot 1024 = 15360 + 5120 = 20480$.

Class 11 Mathematics Binomial Theorem Properties of binomial coefficients

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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 $S_{11} =$

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$18944$
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$17920$
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$20480$
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$40960$

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The sum of the series $aC_0 + (a + b)C_1 + (a + 2b)C_2 + \dots + (a + nb)C_n$ is,where $C_r$ denotes the combinatorial coefficient in the expansion of $(1 + x)^n, n \in N$.

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If the coefficients of the $5^{th}$,$6^{th}$,and $7^{th}$ terms in the expansion of $(1 + x)^n$ are in $A.P.$,then $n =$

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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 $S_{11} =$

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The sum of the series $aC_0 + (a + b)C_1 + (a + 2b)C_2 + \dots + (a + nb)C_n$ is,where $C_r$ denotes the combinatorial coefficient in the expansion of $(1 + x)^n, n \in N$.

If ${}^n C_0, {}^n C_1, {}^n C_2, \ldots, {}^n C_n$ are the binomial coefficients in the expansion of $(1+x)^n$,then for $n=10$,the value of $\sum_{r=1}^{10} {}^n C_r \cdot r(r-4)$ is:

If $\sum\limits_{k=1}^{31} \binom{31}{k} \binom{31}{k-1} - \sum\limits_{k=1}^{30} \binom{30}{k} \binom{30}{k-1} = \frac{\alpha(60!)}{(30!)(31!)}$,where $\alpha \in R$,then the value of $16\alpha$ is equal to

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