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Find the 20th term of (x+3)^22.
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I'd like to also know the process of getting the answer :)
do you have an idea on how to solve it? any advice would help c:
Formula for the k+1 term is \[\left(\begin{matrix}n\\ k\end{matrix}\right)a ^{n-k}b ^{k}=\left(\begin{matrix}22 \\19\end{matrix}\right)x ^{3}5^{19}\]
Wait wait, so I solve that?
Well, what you really need to do is simplify....
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Simplify... how?
Calculate \[\left(\begin{matrix}22 \\ 19\end{matrix}\right)=22!/(19!3!)\] then calculate \[5^{19}\] and multiply. You'll get (REALLY BIG NUMBER)x^5.
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