Chapter 10.2, Problem 46E

To determine

To recognize: The result of the given polynomial as Binomial Theorem.

Explanation of Solution

Giveninformation:

The function

  $f(x)={(1+x)}^m$

Calculation:

The given function is $f(x)={(1+x)}^m$ , m is the positive integer then the McLaurin series generated by $f(x)$ is a polynomial of degree m because $f(x)={(1+x)}^m$ m is a polynomial of degree m. On expanding ${(1+x)}^m$ m then the series of degree m which itself is a McLaurinseries . Alternatively, observe that $f^{(0)}(0)=0$ for $k\ge m+1$

According to binomial theorem

  ${(y+x)}^m=\left(\begin{array}{c}m\\ 0\end{array}\right)y^m{x}^0+\left(\begin{array}{c}m\\ 1\end{array}\right)y^{m-1}{x}^1+\left(\begin{array}{c}m\\ 2\end{array}\right)y^{m-2}{x}^2+\dots \dots \dots +\left(\begin{array}{c}m\\ m-1\end{array}\right)y^1{x}^{m-1}+\left(\begin{array}{c}m\\ m\end{array}\right)y^0{x}^m$ this can be also written as,

  ${(y+x)}^m={\displaystyle \sum }_{k=0}^m\left(\begin{array}{l}m\hfill \\ k\hfill \end{array}\right)y^{m-k}{x}^k$ putting $y=1$ in the above expression.

  ${(1+x)}^m={\displaystyle \sum }_{k=0}^m\left(\begin{array}{l}m\hfill \\ k\hfill \end{array}\right)x^k$

Since, McLaurin series of a function $f(x)={(1+x)}^m$ m can also be represented as,

  ${(1+x)}^m={\displaystyle \sum }_{k=0}^{\infty }\left(\begin{array}{l}m\hfill \\ k\hfill \end{array}\right)x^k$ since, $k\ge m+1$ from the above result .

Therefore, ${(1+x)}^m={\displaystyle \sum }_{k=0}^m\left(\begin{array}{l}m\hfill \\ k\hfill \end{array}\right)x^k$ this result same as the binomial theorem.