91Ó°ÊÓ

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.

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 $\underset{k=0}{\overset{∞}{∑}}\left(\begin{array}{c}p\\ k\end{array}\right)x^k$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.$

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

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

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,

$$\begin{gathered} 1+3x+\frac{3×2}{2!}{x}^2+\frac{3×2×1}{3!}{x}^3 \\ ⇒1+3x+3x^2+x^3 \end{gathered}$$

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.