`(3a - 4b)^5` Use the Binomial Theorem to expand and simplify the expression.

Tuesday, August 15, 2017

$\displaystyle{{\left({3}{a}-{4}{b}\right)}}^{{5}}$ 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 3a for x, 4b for y and 5 for n, such that:

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

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 $\displaystyle{5}{C}{2}={5}{C}{3}={10}.$

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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 3a for x, 4b for y and 5 for n, such that:

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 $\displaystyle{5}{C}{2}={5}{C}{3}={10}.$

$\displaystyle{{\left({3}{a}-{4}{b}\right)}}^{{5}}={243}{{a}}^{{5}}-{1620}{{a}}^{{4}}\cdot{b}+{4320}{{a}}^{{3}}\cdot{{b}}^{{2}}-{5760}{{a}}^{{2}}\cdot{{b}}^{{3}}+{3840}{a}\cdot{{b}}^{{4}}-{1024}{{b}}^{{5}}$

Hence, expanding the complex number using binomial theorem yields the simplified result $\displaystyle{{\left({3}{a}-{4}{b}\right)}}^{{5}}={243}{{a}}^{{5}}-{1620}{{a}}^{{4}}\cdot{b}+{4320}{{a}}^{{3}}\cdot{{b}}^{{2}}-{5760}{{a}}^{{2}}\cdot{{b}}^{{3}}+{3840}{a}\cdot{{b}}^{{4}}-{1024}{{b}}^{{5}}.$