Deriving Maclaurin series

<p>Prove that ${{\left( 1+x \right)}^{n}}=1+nx+\frac{n\left( n-1 \right)}{2!}{{x}^{2}}+\frac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+...$.</p>

Prove that ${{\left( 1+x \right)}^{n}}=1+nx+\frac{n\left( n-1 \right)}{2!}{{x}^{2}}+\frac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+...$.

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**Deriving the Binomial Expansion using Maclaurin Series:**

Let $\mathrm{f}(x) = (1+x)^n$

$\mathrm{f}(x) = (1+x)^n \longrightarrow f(0) = (1+0)^n = 1$
$\mathrm{f}'(x) = n(1+x)^{n-1} \longrightarrow f'(0) = n(1+0)^{n-1} = n$
$\mathrm{f}''(x) = n(n-1)(1+x)^{n-2} \longrightarrow f''(0) = n(n-1)(1+0)^{n-2} = n(n-1)$
$f'''(x) = n(n-1)(n-2)(1+x)^{n-3} \longrightarrow f'''(0) = n(n-1)(n-2)$

From MF26 (Maclaurin series):
$\mathrm{f}(x) = f(0) + xf'(0) + \displaystyle\frac{x^2}{2!}f''(0) + ... + \frac{x^n}{n!}f^{(n)}(0) + ...$

$\therefore (1+x)^n = 1 + x(n) + \displaystyle\frac{x^2}{2!}(n(n-1)) + \frac{x^3}{3!}(n(n-1)(n-2)) + ...$

$= 1 + nx + \displaystyle\frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...$

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Worked Example 1

Worked Example 1

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