Expand $$(1 + x + x^2)(1 − x)^8$$ in ascending powers of $x$, up to and including, $x^3$ using binomial theorem.
I have been trying to study this, and have been using this URL: http://www.purplemath.com/modules/binomial2.htm
Although I understand the fundamentals (not committed to memory 100% but close) for one set of brackets, I do not know how to approach the situation with 2 sets of brackets as shown above.
Please could someone explain the method for this with 2 sets of brackets in a step by step way that is easy to understand, for someone who is pretty much a beginner with this stuff. please keep it within my realm of understanding. I would be very grateful if someone could do this.
edit..
I know that if it was just $(1-8)^8$
then I could plug it into the formula same as below.
(Am i correct to say I would only go until $(1-8)^{8-3}$ for my question ?)
But i DO NOT know how to plug it in to the formula when there are 2 sets of parentheses. my question is how to I apply the formula shown in the picture when there are two parentheses?
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\pars{1 + x + x^{2}}\pars{1 − x}^{8} = \pars{1 - x^{3}}\pars{1 - x}^{7} = \sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n} - \sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n + 3} \\[5mm] = &\ \sum_{n = 0}^{7}{7 \choose n}\pars{-1}^{n}x^{n} + \sum_{n = 3}^{10}{7 \choose n - 3}\pars{-1}^{n}x^{n} \\[5mm] = &\ \bracks{1 - 7x + 21x^{2} + \sum_{n = 3}^{7}{7 \choose n}\pars{-1}^{n}x^{n}} + \bracks{\sum_{n = 3}^{7}{7 \choose n - 3}\pars{-1}^{n}x^{n} + 21x^{8} - 7x^{9} + x^{10}} \\[5mm] = &\ 1 - 7x + 21x^{2} + \sum_{n = 3}^{7}\bracks{{7 \choose n} + {7 \choose n - 3}}\pars{-1}^{n}x^{n} + 21x^{8} - 7x^{9} + x^{10} = \sum_{n = 0}^{10}a_{n}x^{n} \end{align} where $$ a_{n} = \left\{\begin{array}{lcl} \ds{\phantom{-}1} & \mbox{if} & \ds{n = 0} \\[2mm] \ds{-7} & \mbox{if} & \ds{n = 1} \\[2mm] \ds{\phantom{-}21} & \mbox{if} & \ds{n = 2} \\[2mm] \ds{\pars{-1}^{n}\bracks{% {7 \choose n} + {7 \choose n - 3}}} & \mbox{if} & \ds{3 \leq n \leq 7} \\[2mm] \ds{a_{10 -n}} & \mbox{if} & \ds{8 \leq n \leq 10} \end{array}\right. $$ Indeed, the coefficients $\ds{\braces{a_{n}}}_{\ 0\ \leq\ n\ \leq\ 10}$ satisfacen $\ds{a_{n} = a_{10 - n}}$ such that's sufficient to enumerate the values of $\ds{a_{n}}$ for $\ds{n = 0,1,\ldots,5}$. Namely, $$ a_{0} = 1\,,\ a_{1} = -7\,,\ a_{2} = 21\,, a_{3} = -36\,,\ a_{4} = 42\,,\ a_{5} = -42 $$