(A) The given series is $S = \sum_{r=0}^{n} (3 + 4r) C_r$. We know that $C_r = \binom{n}{r}$. So,$S = 3 \sum_{r=0}^{n} C_r + 4 \sum_{r=0}^{n} r C_r$.

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(A) The given series is $S = \sum_{r=0}^{n} (3 + 4r) C_r$. We know that $C_r = \binom{n}{r}$. So,$S = 3 \sum_{r=0}^{n} C_r + 4 \sum_{r=0}^{n} r C_r$. 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,we get $S = 3(2^n) + 4(n 2^{n-1})$. $S = 3 \cdot 2^n + 2n \cdot 2^n = (2n + 3) 2^n$.

Class 11 Mathematics Binomial Theorem Properties of binomial coefficients

Medium

$3 \cdot C_0 + 7 \cdot C_1 + 11 \cdot C_2 + \ldots + (3 + 4n) C_n =$

AP EAMCET 2017 All AP EAMCET

A
$(2n + 3) 2^n$
B
$(2n + 1) 2^{n-1}$
C
$(2n + 3) 2^{n-1}$
D
$(2n + 1) 2^n$

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$3 \cdot C_0 + 7 \cdot C_1 + 11 \cdot C_2 + \ldots + (3 + 4n) C_n =$

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