If {}^{21}C_1 + 3 cdot {}^{21}C_3 + 5 cdot {} | vedclass

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(C) The given sum is $S = \sum_{r=0}^{10} (2r+1) \cdot {}^{21}C_{2r+1}$.Using the identity $n \cdot {}^{n-1}C_{r-1} = r \cdot {}^{n}C_r$,we have $r \c

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$3$

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(C) The given sum is $S = \sum_{r=0}^{10} (2r+1) \cdot {}^{21}C_{2r+1}$.Using the identity $n \cdot {}^{n-1}C_{r-1} = r \cdot {}^{n}C_r$,we have $r \cdot {}^{n}C_r = n \cdot {}^{n-1}C_{r-1}$.Thus,the sum is $\sum_{r=0}^{10} (2r+1) \cdot {}^{21}C_{2r+1} = \sum_{r=0}^{10} 21 \cdot {}^{20}C_{2r} = 21 \cdot ( {}^{20}C_0 + {}^{20}C_2 + \dots + {}^{20}C_{20} ).$We know that $\sum_{k 	ext{ even}} {}^{n}C_k = 2^{n-1}$.So,$k = 21 \cdot 2^{20-1} = 21 \cdot 2^{19} = 3 \cdot 7 \cdot 2^{19}$.The prime factors of $k$ are $2, 3,$ and $7$.Therefore,the number of prime factors is $3$.

List-$I$	List-$II$
$(A)$ $a_0 + a_2 + \ldots + a_{2n}$	$(I)$ $n \cdot 3^{n-1}$
$(B)$ $a_1 + a_3 + \ldots + a_{2n-1}$	$(II)$ $n \cdot 3^n$
$(C)$ $a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$	$(III)$ $\frac{1}{2}(3^n + 1)$
	$(IV)$ $\frac{1}{2}(3^n - 1)$

If ${}^{21}C_1 + 3 \cdot {}^{21}C_3 + 5 \cdot {}^{21}C_5 + \dots + 19 \cdot {}^{21}C_{19} + 21 \cdot {}^{21}C_{21} = k$,then the number of prime factors of $k$ is:

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