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The binomial series expansion for an expression of the form \$$(a+bx)^n\$$ where \$$n\$$ is a Natural number is given by:

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\$$(a+bx)^n=\\tbinom{n}{0}(a)^n(bx)^{0}+\\tbinom{n}{1}(a)^{n-1}(bx)^{1}+\\tbinom{n}{2}(a)^{n-2}(bx)^{2}+...+\\tbinom{n}{k}(a)^{n-k}(bx)^{k}+...+\\tbinom{n}{n}(a)^{0}(bx)^{n}\$$

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In this example  \$$n=\\var{n}\$$,  \$$a=\\var{a}\$$  and  \$$b=\\var{b}\$$.

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So the binomial series expansion is:

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\$$\\var{a}^{\\var{n}}+\\tbinom{\\var{n}}{\\var{1}}\\times\\var{a}^{\\var{n}-1}\\times(\\var{b}x)^{1}+\\tbinom{\\var{n}}{2}\\times\\var{a}^{\\var{n}-2}\\times(\\var{b}x)^{2}+\\var{ellipsis}\\tbinom{\\var{n}}{\\var{n}}\\times\\var{a}^{\\var{n}-\\var{n}}\\times(\\var{b}x)^\\var{n}\$$

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\$$=\\simplify{{a}^{n}+{n}*{a}^({n}-1)*{b}x+{n}*{n-1}*{a}^{{n}-2}*{{b}^2}/2x^2+{n}*{n-1}*{n-2}*{a}^{{n}-3}*{{b}^3}/6x^3+{n}*{n-1}*{n-2}*{n-3}*{a}^{{n}-4}*{{b}^4}/24x^4+{n}*{n-1}*{n-2}*{n-3}*{n-4}*{a}^{{n}-5}*{{b}^5}/120x^5}\$$

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Given the expression $\\simplify{({a}+{b}x)^{n}}$