Let S_1 = sum_{j=1}^{10} j(j-1) binom{10}{j}, | vedclass

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(C) We know that $\sum_{j=0}^{n} \binom{n}{j} x^j = (1+x)^n$. For $n=10$,$\sum_{j=0}^{10} \binom{10}{j} x^j = (1+x)^{10}$. Differentiating with respec

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$(A)$ is true,but $(R)$ is false

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(C) We know that $\sum_{j=0}^{n} \binom{n}{j} x^j = (1+x)^n$. For $n=10$,$\sum_{j=0}^{10} \binom{10}{j} x^j = (1+x)^{10}$. Differentiating with respect to $x$: $\sum_{j=1}^{10} j \binom{10}{j} x^{j-1} = 10(1+x)^9$. Setting $x=1$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j} = 10(2^9) = 5 	imes 2^{10} = 20 	imes 2^8$. Thus,the value of $S_2$ in Reason $(R)$ is incorrect. For $S_1$,$\sum_{j=2}^{10} j(j-1) \binom{10}{j} x^{j-2} = 10 	imes 9(1+x)^8$. Setting $x=1$,$S_1 = 90 	imes 2^8$. Since $S_3 = \sum j^2 \binom{10}{j} = \sum (j(j-1) + j) \binom{10}{j} = S_1 + S_2 = 90 	imes 2^8 + 10 	imes 2^9 = 90 	imes 2^8 + 20 	imes 2^8 = 110 	imes 2^8 = 55 	imes 2^9$. Assertion $(A)$ is true,but Reason $(R)$ is false because $S_2 = 20 	imes 2^8$.

Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Assertion $(A) : S_3 = 55 \times 2^9$
Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$

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Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.Assertion $(A) : S_3 = 55 \times 2^9$Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$