Expand (1+x+x2)(1âˆ′x)8(1 + x + x^2)(1 âˆ’ x)^8 in ascending powers of xx, up to and including, x3x^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(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(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?

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2

Could you please expand this as far as you are able? Or expand one of the terms with binomial theorem.

– Larry B.

Oct 20 at 22:39

So I guess you started by using the binomial theorem to expand (1−x)8(1-x)^8 and then you multiplied the resulting expansion by 1+x+x2?1+x+x^2? At what point did you get stuck?

– bof

Oct 20 at 22:42

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1 Answer

1

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\begin{align}

&\pars{1 + x + x^{2}}\pars{1 âˆ’ 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}\begin{align}

&\pars{1 + x + x^{2}}\pars{1 âˆ’ 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.

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}\ds{\braces{a_{n}}}_{\ 0\ \leq\ n\ \leq\ 10} satisfacen \ds{a_{n} = a_{10 – n}}\ds{a_{n} = a_{10 – n}} such that’s sufficient to enumerate the values of \ds{a_{n}}\ds{a_{n}} for \ds{n = 0,1,\ldots,5}\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

a_{0} = 1\,,\ a_{1} = -7\,,\ a_{2} = 21\,, a_{3} = -36\,,\ a_{4} = 42\,,\

a_{5} = -42

I appreciate you putting time into this, please can you read my edit as it shows the method that I am beginning to understand, this topic is very confusing to me and the answers have also confused me. can you please explain in a way that I can see visually connects to the image I have added ?

– Flewitt Connor

2 days ago