Question

Show that if $p$is a positive integer, then the binomial series for $f\left(x\right)={(1+x)}^p$is a polynomial.

Short answer

Hence, we have shown that if$p$is a positive integer, then the binomial series for$f\left(x\right)={(1+x)}^p$is a polynomial.

Step-by-step solution

Step 1. Given information

We have been asked to prove that if $p$is a positive integer, then the binomial series for $f\left(x\right)={(1+x)}^p$is a polynomial.

Step 2. Defining the series

Given that $f\left(x\right)={(1+x)}^p$.

For any non-zero constant$p$, the Maclaurin series for the function $f\left(x\right)={(1+x)}^p$is called the binomial series which is given by $\sum _{k=0}^{\infty }\left(\begin{array}{c}p\\ k\end{array}\right)x^k$where the binomial coefficient is$\left(\begin{array}{c}p\\ k\end{array}\right)$$\left(\begin{array}{c}p\\ k\end{array}\right)=\left\{\begin{array}{l}\frac{p(p-1)(p-2)...(p-k+1)}{k!},ifk>0\\ 1,ifk=0\end{array}\right.$

Step 3. Binomial series of f(x)=(1+x)p

Therefore, the binomial series of $f\left(x\right)={(1+x)}^p$is

$1+px+\frac{p(p-1)}{2!}{x}^2+...+\frac{p(p-1)(p-2)...(p-k+1)}{n!}{x}^k+...$

Step 4. Proof

Since $p$is a positive integer, let us take $p=3$,

Then the binomial series for the function $f\left(x\right)={(1+x)}^p$
is,

$$\begin{gathered} 1+3x+\frac{3\times 2}{2!}{x}^2+\frac{3\times 2\times 1}{3!}{x}^3 \\ \Rightarrow 1+3x+3x^2+x^3 \end{gathered}$$

which is a polynomial.

Hence, we have proved that if $p$is a positive integer, then the binomial series forlocalid="1649601815534" $f\left(x\right)={(1+x)}^p$is a polynomial.