Notes: `(x + y)^5` Use the Binomial Theorem to expand and simplify the expression.

Monday, December 21, 2015

$\displaystyle{{\left({x}+{y}\right)}}^{{5}}$ Use the Binomial Theorem to expand and simplify the expression.

You need to use the binomial formula, such that:

$\displaystyle{{\left({x}+{y}\right)}}^{{n}}={\sum_{{{k}={0}}}^{{n}}}{\left(\matrix{{n}\\{k}}\right)}{{x}}^{{{n}-{k}}}{{y}}^{{k}}$

You need to replace x for x, y for y and 5 for n, such that:

$\displaystyle{{\left({x}+{y}\right)}}^{{5}}={5}{C}{0}{{\left({x}\right)}}^{{5}}+{5}{C}{1}{{\left({x}\right)}}^{{4}}\cdot{{\left({y}\right)}}^{{1}}+{5}{C}{2}{{\left({x}\right)}}^{{3}}\cdot{{\left({y}\right)}}^{{2}}+{5}{C}{3}{{x}}^{{2}}\cdot{{y}}^{{3}}+{5}{C}{4}{x}\cdot{{y}}^{{4}}+{5}{C}{5}{{y}}^{{5}}$

By definition, nC0 = nCn = 1, hence $\displaystyle{5}{C}{0}={5}{C}{5}={1}.$

By definition nC1 = nC(n-1) = n, hence $\displaystyle{5}{C}{1}={5}{C}{4}={5}.$

By definition $\displaystyle{n}{C}{2}=\frac{{{n}{\left({n}-{1}\right)}}}{{2}}$ , hence $\displaystyle{5}{C}{2}={5}{C}{3}={10}.$

$\displaystyle{{\left({x}+{y}\right)}}^{{5}}={{x}}^{{5}}+{5}{{x}}^{{4}}\cdot{y}+{10}{{x}}^{{3}}\cdot{{y}}^{{2}}+{10}{{x}}^{{2}}\cdot{{y}}^{{3}}+{5}{x}\cdot{{y}}^{{4}}+{{y}}^{{5}}$

Hence, expanding the complex number using binomial theorem yields the simplified result $\displaystyle{{\left({x}+{y}\right)}}^{{5}}={{x}}^{{5}}+{5}{{x}}^{{4}}\cdot{y}+{10}{{x}}^{{3}}\cdot{{y}}^{{2}}+{10}{{x}}^{{2}}\cdot{{y}}^{{3}}+{5}{x}\cdot{{y}}^{{4}}+{{y}}^{{5}}.$

Notes

Monday, December 21, 2015

$\displaystyle{{\left({x}+{y}\right)}}^{{5}}$ Use the Binomial Theorem to expand and simplify the expression.

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