Thread: please find coefficient

nb

Originally Posted by oceanmd

I do not even know how to touch this problem. Please help.
Find coefficient of x^0 in the expansion of (x^2+1\x)^12

Thank you

we have $\displaystyle \left( x^2 + x^{-1} \right)^{12}$

using $\displaystyle (a + b)^n = \sum_{k = 0}^{n}{n \choose k}a^kb^{n - k}$ we have:

$\displaystyle \left( x^2 + \frac 1x \right)^{12} = \left( x^2 + x^{-1} \right)^{12} = \sum_{k = 0}^{12}{12 \choose k}x^{2k}x^{k - 12} = \sum_{k = 0}^{12}{12 \choose k}x^{3k - 12}$

now continue

Originally Posted by oceanmd

3k-12=0; k=4
coefficient is 495.
Thank you

yes