(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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Binomial Theorem

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If for $3 \leq r \leq 30$, $\binom{30}{30-r} + 3\binom{30}{31-r} + 3\binom{30}{32-r} + \binom{30}{33-r} = \binom{m}{r}$, then $m$ equals:

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If $(1+x)^n = p_0 + p_1 x + p_2 x^2 + \ldots + p_n x^n$,then the value of $p_0 + p_3 + p_6 + \ldots$ is equal to:

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In the expansion of $(1 + x)^n$,the sum of the coefficients of odd powers of $x$ is:

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The sum of the last eight coefficients in the expansion of $(1 + x)^{15}$ is

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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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If for $3 \leq r \leq 30$, $\binom{30}{30-r} + 3\binom{30}{31-r} + 3\binom{30}{32-r} + \binom{30}{33-r} = \binom{m}{r}$, then $m$ equals:

If $(1+x)^n = p_0 + p_1 x + p_2 x^2 + \ldots + p_n x^n$,then the value of $p_0 + p_3 + p_6 + \ldots$ is equal to:

In the expansion of $(1 + x)^n$,the sum of the coefficients of odd powers of $x$ is:

The sum of the last eight consecutive coefficients in the expansion of $(1+x)^{15}$ is

The sum of the last eight coefficients in the expansion of $(1 + x)^{15}$ is

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