Could anyone help me solve me the following problem please :-
Find the coefficent of x^27 in the binomial expansion of ( 3/x + x^2 )^18.
I can work my way through the initial steps but do not know how to go on from there. Thanks in advance.
Could anyone help me solve me the following problem please :-
Find the coefficent of x^27 in the binomial expansion of ( 3/x + x^2 )^18.
I can work my way through the initial steps but do not know how to go on from there. Thanks in advance.
By Newton's binomial Theorem, $\displaystyle \left(\frac{3}{x}+x^2\right)^{18}=\sum\limits_{k=0 }^{18}\binom {18}{k}3^kx^{-k}x^{2(18-k)}$ $\displaystyle =\sum\limits_{k=0}^{18}\binom {18}{k}3^kx^{36-3k}$
Since you want the coefficient of $\displaystyle x^{27}$ , you need to know for which index k we get
$\displaystyle 36-3k=27\Longleftrightarrow 3k=9\Longleftrightarrow k=3$ , so the wanted coefficient is $\displaystyle \binom{18}{3}3^3$.
Now do the maths and evaluate this number.
tonio
Hello, aamiri!
$\displaystyle \text{Find the coefficent of }x^{27}\text{ in the binomial expansion of }\left(\frac{3}{x} + x^2 \right)^{18}$
We have: .$\displaystyle \left(3x^{-1} + x^2\right)^{18}$
The terms are of the form: .$\displaystyle \displaystyle {18\choose a,b}\left(3x^{-1}\right)^a\left(x^2\right)^b \;\;\text{ where }a+b \,=\,18$ .[1]
The terms we want are of the form: .$\displaystyle \left(x^{-1}\right)^a(x^2)^b \;=\;x^{27}$
. . Then: .$\displaystyle (x^{-a})(x^{2b}) \:=\:x^{27} \quad\Rightarrow \quad x^{2b-a} \:=\:x^{27}$
$\displaystyle \begin{array}{ccccc}\text{We have:} & 2b - a &=& 27 \\
\text{Add [1]:} & b + a &=& 18 \end{array}$
Hence: .$\displaystyle 3b \,=\,45 \quad\Rightarrow\quad b\,=\,15 \quad\Rightarrow\quad a\,=\,3$
It seems that there is only one term that contains $\displaystyle x^{27}.$
That term is: .$\displaystyle \displaystyle {18\choose3,15}\left(3x^{-1}\right)^3\left(x^2\right)^{15}$
. . . . . . . . $\displaystyle =\;(816)\,\left(3^3x^{-3}\right)\left(x^{30}\right) $
. . . . . . . . $\displaystyle =\;\boxed{22,\!032}\,x^{27}$
Edit: Ha! . . . This is rehash of tonio's solution . . . *blush*