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

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Question

Instead of ${ }^{n} C_{k}$ it must be ${ }^{10} C_{k}$ i.e.

$\sum_{k=0}^{10}\left(2^{2}+3 k\right){ }^{10} C _{ k }=\alpha .3^{10}+\beta .2^{10}$

Vedclass · Accepted Answer

$19$

Vedclass · Answer

Instead of ${ }^{n} C_{k}$ it must be ${ }^{10} C_{k}$ i.e.

$\sum_{k=0}^{10}\left(2^{2}+3 k\right){ }^{10} C _{ k }=\alpha .3^{10}+\beta .2^{10}$

$LHS =4 \sum_{ k =0}^{10}{ }^{10} C _{ k }+3 \sum_{ k =0}^{10} k \cdot \frac{10}{ k } \cdot{ }^{9} C _{ k -1}$

$=4.2^{10}+3.10 .2^{9}$

$=19.2^{10}=\alpha .3^{10}+\beta .2^{10}$

$\Rightarrow \alpha=0, \beta=19 \Rightarrow \alpha+\beta=19$

Let ${ }^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+ x )^{ n }.$

If $\sum_{ k =0}^{10}\left(2^{2}+3 k \right){ }^{ n } C _{ k }=\alpha .3^{10}+\beta \cdot 2^{10}, \alpha, \beta \in R$ then $\alpha+\beta$ is equal to ....... .

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