# Express the following sum in closed form (without using a summation symbol and without using...

## Question:

Express the following sum in closed form (without using a summation symbol and without using ellipsis ...) :

{eq}\sum\limits_{k=0}^n(^n_k)9^k {/eq}

Explain all steps

## Binomial Expansion:

{eq}\\ {/eq}

The binomial expansion of the expression: {eq}(a+b)^n {/eq} is:

{eq}(a+b)^n = \sum_{k=0}^{n} a^k \ b^{n-k} \ {n\choose k} {/eq}

The binomial coefficient is defined as: {eq}{n\choose k} = \frac{n!}{k!(n-k)!} {/eq}

{eq}\\ {/eq}

The binomial expansion of {eq}(a+b)^n {/eq} is:

{eq}\begin{align*} (a+b)^n &= \sum_{k=0}^{n} a^k \ b^{n-k} \ {n\choose k} &&(1) \end{align*} {/eq}

Let {eq}S {/eq} be the closed form sum of the given series.

We have:

{eq}\begin{align*} S &= \sum_{k=0}^{n} 9^k {n\choose k} \\ &= \sum_{k=0}^{n} 9^k \ 1^{n-k} \ {n\choose k} &&[\because 1^{n-k} = 1 ] \\ &= (9+1)^n &&[\text{compare with (1) }: a = 9, \ b=1] \\ \therefore \ \sum_{k=0}^{n} 9^k {n\choose k} &= 10^n \end{align*} {/eq}

{eq}{/eq}