`(x + 2y)^4` Use the Binomial Theorem to expand and simplify the expression.

Friday, May 20, 2016

$\displaystyle{{\left({x}+{2}{y}\right)}}^{{4}}$ 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, 2y for y and 4 for n, such that:

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

By definition, $\displaystyle{n}{C}{0}={n}{C}{n}={1}$ , hence $\displaystyle{4}{C}{0}={4}{C}{4}={1}$ .

By definition $\displaystyle{n}{C}{1}={n}{C}{\left({n}-{1}\right)}={n}$ , hence $\displaystyle{4}{C}{1}={4}{C}{3}={4}.$

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

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, 2y for y and 4 for n, such that:

By definition, $\displaystyle{n}{C}{0}={n}{C}{n}={1}$ , hence $\displaystyle{4}{C}{0}={4}{C}{4}={1}$ .

By definition $\displaystyle{n}{C}{1}={n}{C}{\left({n}-{1}\right)}={n}$ , hence $\displaystyle{4}{C}{1}={4}{C}{3}={4}.$

By definition $\displaystyle{n}{C}{2}={n}{C}{\left({n}-{2}\right)}=\frac{{{n}{\left({n}-{1}\right)}}}{{2}}$ , hence $\displaystyle{4}{C}{2}=\frac{{{4}{\left({4}-{1}\right)}}}{{2}}={6}$

$\displaystyle{{\left({x}+{2}{y}\right)}}^{{4}}={{x}}^{{4}}+{4}{{x}}^{{3}}\cdot{{\left({2}{y}\right)}}^{{1}}+{6}{{x}}^{{2}}\cdot{4}{{y}}^{{2}}+{4}{x}\cdot{8}{{y}}^{{3}}+{16}{{y}}^{{4}}$

$\displaystyle{{\left({x}+{2}{y}\right)}}^{{4}}={{x}}^{{4}}+{8}{{x}}^{{3}}\cdot{y}+{24}{{x}}^{{2}}\cdot{{y}}^{{2}}+{32}{x}\cdot{{y}}^{{3}}+{16}{{y}}^{{4}}$

Hence, expanding the complex number using binomial theorem yields the simplified result $\displaystyle{{\left({x}+{2}{y}\right)}}^{{4}}={{x}}^{{4}}+{8}{{x}}^{{3}}\cdot{y}+{24}{{x}}^{{2}}\cdot{{y}}^{{2}}+{32}{x}\cdot{{y}}^{{3}}+{16}{{y}}^{{4}}.$

at May 20, 2016

Notes

Friday, May 20, 2016

$\displaystyle{{\left({x}+{2}{y}\right)}}^{{4}}$ Use the Binomial Theorem to expand and simplify the expression.

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Friday, May 20, 2016

Use the Binomial Theorem to expand and simplify the expression.

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$\displaystyle{{\left({x}+{2}{y}\right)}}^{{4}}$ Use the Binomial Theorem to expand and simplify the expression.