Let { }^{n} C_{r} denote the binomial coeffic | vedclass

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(A) Given the sum $\sum_{k=0}^{10}(4+3k){ }^{10} C _{ k } = \alpha \cdot 3^{10} + \beta \cdot 2^{10}.$We know that $\sum_{k=0}^{n} {}^{n}C_k = 2^n$ an

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$19$

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(A) Given the sum $\sum_{k=0}^{10}(4+3k){ }^{10} C _{ k } = \alpha \cdot 3^{10} + \beta \cdot 2^{10}.$We know that $\sum_{k=0}^{n} {}^{n}C_k = 2^n$ and $\sum_{k=0}^{n} k \cdot {}^{n}C_k = n \cdot 2^{n-1}.$Expanding the sum: $\sum_{k=0}^{10} 4 \cdot {}^{10}C_k + 3 \sum_{k=0}^{10} k \cdot {}^{10}C_k.$$= 4 \cdot 2^{10} + 3 \cdot (10 \cdot 2^{9}).$$= 4 \cdot 2^{10} + 30 \cdot 2^{9} = 4 \cdot 2^{10} + 15 \cdot 2^{10} = 19 \cdot 2^{10}.$Comparing with $\alpha \cdot 3^{10} + \beta \cdot 2^{10},$ we get $\alpha = 0$ and $\beta = 19.$Therefore,$\alpha + \beta = 0 + 19 = 19.$

Let ${ }^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+ x )^{ n }.$ If $\sum_{ k =0}^{10}\left(2^{2}+3 k \right){ }^{10} C _{ k }=\alpha \cdot 3^{10}+\beta \cdot 2^{10},$ where $\alpha, \beta \in R,$ then $\alpha+\beta$ is equal to ....... .

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