Question

Let ${}^{n}C_r$ denote the binomial coefficient of $x^r$ in the expansion of ${(1+x)}^n$. If $\sum _{k=0}^{10}\left[2^2+3k\right]{}^{n}C_k=\alpha .3^{10}+\beta .2^{10}$ then $ɑ+\beta$ is equal to

Answer 1

Finding the value of $\alpha +\beta$:

Given, $\sum _{k=0}^{10}\left[2^2+3k\right]{}^{n}C_k=\alpha .3^{10}+\beta .2^{10}$

We know, ${\left(1+x\right)}^n=\sum _{r=0}^n{}^{n}C_r{x}^r$

Now, simplifying the above expression we get,

$$\begin{gathered} \sum _{k=0}^{10}\left[2^2+3k\right]{}^{n}C_k=4\sum _{k=0}^{10}{}^{n}C_k+3\sum _{k=0}^{10}k{}^{n}C_k \\ =4{\left(2\right)}^{10}+3\times \sum _{k=0+1}^{10}k\times \frac{10}{k}\times {}^{9}C_{k-1}\left[\text{Here},n=10,x=1,{}^{n}C_r=\frac{n}{r}\times {}^{n-1}C_{r-1}\right] \\ =4{\left(2\right)}^{10}+(3\times 10\times 2^9) \\ =4{\left(2\right)}^{10}+3\times 5\times 2^{10} \\ =4{\left(2\right)}^{10}+15{\left(2\right)}^{10} \\ =19{\left(2\right)}^{10} \end{gathered}$$

Comparing with the given equation $\alpha .3^{10}+\beta .2^{10}$ , we get $\alpha =0,\beta =19$

Therefore, $\alpha +\beta =19$ is the value of given expression.