`(x + 1)^6` Use the Binomial Theorem to expand and simplify the expression.

Friday, January 24, 2014

$\displaystyle{{\left({x}+{1}\right)}}^{{6}}$ Use the Binomial Theorem to expand and simplify the expression.

We have to expand the expression using the Binomial Theorem. So use the binomial formula,

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

$\displaystyle\therefore{{\left({x}+{1}\right)}}^{{6}}={\left(\matrix{{6}\\{0}}\right)}{{x}}^{{{6}-{0}}}\cdot{{1}}^{{0}}+{\left(\matrix{{6}\\{1}}\right)}{{x}}^{{{6}-{1}}}\cdot{{1}}^{{{1}}}+{\left(\matrix{{6}\\{2}}\right)}{{x}}^{{{6}-{2}}}\cdot{{1}}^{{2}}+{\left(\matrix{{6}\\{3}}\right)}{{x}}^{{{6}-{3}}}\cdot{{1}}^{{3}}+{\left(\matrix{{6}\\{4}}\right)}{{x}}^{{{6}-{4}}}\cdot{{1}}^{{4}}+{\left(\matrix{{6}\\{5}}\right)}{{x}}^{{{6}-{5}}}\cdot{{1}}^{{5}}+{\left(\matrix{{6}\\{6}}\right)}{{x}}^{{{6}-{6}}}\cdot{{1}}^{{6}}$

$\displaystyle={{x}}^{{6}}+\frac{{{6}!}}{{{1}!{\left({6}-{1}\right)}!}}{{x}}^{{5}}+\frac{{{6}!}}{{{2}!{\left({6}-{2}\right)}!}}{{x}}^{{4}}+\frac{{{6}!}}{{{3}!{\left({6}-{3}\right)}!}}{{x}}^{{3}}+\frac{{{6}!}}{{{4}!{\left({6}-{4}\right)}!}}{{x}}^{{2}}+\frac{{{6}!}}{{{5}!{\left({6}-{5}\right)}!}}{{x}}^{{1}}+{1}$

$\displaystyle={{x}}^{{6}}+\frac{{{6}\cdot{5}!}}{{{5}!}}{{x}}^{{5}}+\frac{{{6}\cdot{5}\cdot{4}!}}{{{2}\cdot{1}\cdot{4}!}}{{x}}^{{4}}+\frac{{{6}\cdot{5}\cdot{4}\cdot{3}!}}{{{3}\cdot{2}\cdot{1}\cdot{3}!}}{{x}}^{{3}}+\frac{{{6}\cdot{5}\cdot{4}!}}{{{4}!\cdot{2}\cdot{1}}}{{x}}^{{2}}+\frac{{{6}\cdot{5}!}}{{{5}!}}{x}+{1}$