(D) The given series is $S = \sum_{r=0}^{n} (a + rb)C_r$.This can be split into two parts:$S = a \sum_{r=0}^{n} C_r + b \sum_{r=0}^{n} r C_r$.We know

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(D) The given series is $S = \sum_{r=0}^{n} (a + rb)C_r$.This can be split into two parts:$S = a \sum_{r=0}^{n} C_r + b \sum_{r=0}^{n} r C_r$.We know that $\sum_{r=0}^{n} C_r = 2^n$ and $\sum_{r=0}^{n} r C_r = n 2^{n-1}$.Substituting these values:$S = a(2^n) + b(n 2^{n-1})$.Factoring out $2^{n-1}$:$S = 2a(2^{n-1}) + bn(2^{n-1})$.$S = (2a + nb) 2^{n-1}$.

Class 11 Mathematics Binomial Theorem Properties of binomial coefficients

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 $(1 + x)^n, n \in N$.

A
$(a + 2nb)2^n$
B
$(2a + nb)2^n$
C
$(a + nb)2^{n - 1}$
D
$(2a + nb)2^{n - 1}$

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Binomial Theorem

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Properties of binomial coefficients

If ${ }^{n} C_0+\frac{1}{2}{ }^{n} C_1+\frac{1}{3}{ }^{n} C_2+\ldots+\frac{1}{n+1}{ }^{n} C_{n}=\frac{1023}{10}$,then $n=$

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If $\sum\limits_{K = 1}^{12} {12K \cdot {^{12}C_K} \cdot {^{11}C_{K - 1}}} $ is equal to $\frac{{12 \times 21 \times 19 \times 17 \times \dots \times 3}}{{11!}} \times {2^{12}} \times p$,then $p$ is:

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If $\sum_{k=1}^{30} k \left({ }^{30} C _k\right)^2 = \frac{\alpha 60 !}{(30 !)^2}$,then $\alpha$ is equal to

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If $\frac{1}{n+1} {}^{n}C_{n} + \frac{1}{n} {}^{n}C_{n-1} + \dots + \frac{1}{2} {}^{n}C_{1} + {}^{n}C_{0} = \frac{1023}{10}$,then $n$ is equal to

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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:

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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 ${ }^{n} C_0+\frac{1}{2}{ }^{n} C_1+\frac{1}{3}{ }^{n} C_2+\ldots+\frac{1}{n+1}{ }^{n} C_{n}=\frac{1023}{10}$,then $n=$

If $\sum\limits_{K = 1}^{12} {12K \cdot {^{12}C_K} \cdot {^{11}C_{K - 1}}} $ is equal to $\frac{{12 \times 21 \times 19 \times 17 \times \dots \times 3}}{{11!}} \times {2^{12}} \times p$,then $p$ is:

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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:

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