Find the coefficient of x^5 in the expansion of (x^2 - 3/x)^7.
Hi, how do you solve this question using binomial theorem. Thanks.
Please help
Ever term in the expansion of $\displaystyle \left( {x^2 - 3x^{ - 1} } \right)^7 $ has the form $\displaystyle C\left( {x^2 } \right)^j \left( { - 3x^{ - 1} } \right)^k ,\;j + k = 7\;\& \quad C = {7 \choose j}$.
$\displaystyle So \left\{ \begin{array}{r}
j + k = 7 \\
2j - k = 5 \\
\end{array} \right.$
For the x's only:
1st term, (x^2)^7 = x^14
2nd term, [(x^2)^6][1/x] = x^11
3rd term, [(x^2)^5][1/(x^2)] = x^8
4th term, [(x^2)^4][1/(x^3)] = x^5 -------so it is in the 4th term.
The whole 4th term has a binomial coefficent of
nCr = 7C4
= 7! / [(7-4)! *4!]
= 7! / [3! *4!]
= 7*6*5 / 3*2*1
= 35
So, the whole 4th term is
35[(x^2)^4][(-3/x)^3]
= 35(-27)x^5
= -945x^5
Therefore, the coefficient is -945. -------answer.