Question

Show that if p is a positive integer, then$\left(\begin{array}{l}p\\ n\end{array}\right)=0$when $n>p$ , so $(1+x{)}^p=\sum \left(\begin{array}{l}p\\ n\end{array}\right)x^n$is just a sum of$p+1$terms, from $n=0$ to $n=p$ . For example,$(1+x{)}^2$has $3$ terms,$(1+x{)}^3$ has$4$terms, etc. This is just the familiar binomial theorem.

Short answer

The statement has been proven.

Step-by-step solution

Given Information

Thebinomial series.

Definition of the binomial series.

The Taylor series for the function given by is the binomial series, where is an arbitrary complex number.

Prove the statement.

The binomial series states that $(1+x{)}^p=\sum _{n=0}^{\infty }\left(\begin{array}{l}p\\ n\end{array}\right)x^n$

The formula states that$\left(\begin{array}{l}p\\ n\end{array}\right)=\frac{p(p-1)(p-2)\dots (p-n+1)}{n!}$

localid="1657347784194" $\left(\begin{array}{l}p\\ n\end{array}\right)=\frac{p(p-1)(p-2)\dots (p-p)\dots (p-n+1)}{n!}$

localid="1657347855207" $\frac{p(p-1)(p-2)\dots (p-p)\dots (p-n+1)}{n!}=0$

localid="1657347906098" $\frac{p!}{p!(p-p)!}=1$

localid="1657347974064" $\left(\begin{array}{l}p\\ n\end{array}\right)\ne 0$ only for localid="1657348003107" $n\le p$andlocalid="1657348034900" $n\ge 0$

Solve further.

localid="1657348111696" $(1+x{)}^p=\sum _{n=0}^{\infty }\left(\begin{array}{l}p\\ n\end{array}\right)x^n$

localid="1657348136816" $(1+x{)}^p=\left(\begin{array}{l}p\\ 0\end{array}\right)+\left(\begin{array}{l}p\\ 1\end{array}\right)x+\left(\begin{array}{l}p\\ 2\end{array}\right)x^2+\left(\begin{array}{l}p\\ 3\end{array}\right)x^3+\dots +\left(\begin{array}{l}p\\ p\end{array}\right)x^p$

The expansion haslocalid="1657348245474" $p+1$terms for localid="1657348169040" $n=0-p$
.

The statement has been proven.