If the middle term of the expansion of (*x*^{3} + 2y^{2})* ^{n}* is

*C*

*x*

^{18}

*y*

^{12}, find

*C*.

April 25, 2017 - 8:04am

#1
Romel

If the middle term of the expansion of (x^3 + 2y^2)^n is Cx^(18) y^(12), find C

If the middle term of the expansion of (*x*^{3} + 2y^{2})* ^{n}* is

SPONSORED LINKS

Here it is:

I let $u = x^3$ and $v = 2y^2.$ Then doing this: $$(u)^6 = (x^3)^6 \space and \space \space \left(\frac{v}{2} \right)^6 = (y^2)^6$$

we get $u^6 = x^{18}$ and $\left(\frac{v^6}{64} \right) = y^{12}$

We now conclude that the expression $(u + v)^n$ has a term $(u^6)\left(\frac{v^6}{64} \right)$ along its expansion when $u = x^3$ and $v = 2y^2$ We need to find its equivalent term of $(u^6)\left(\frac{v^6}{64} \right)$ when we go back to dealing with $(x^3 + 2y^2)^n.$

Everybody knows that in the binomial expansion of $(u + v)^n,$ in each term, the sum of the exponents of $u$ and $v$ is $n$ and there are $n+1$ terms.

With that in mind, the sum of the particular term $(u^6)\left(\frac{v^6}{64} \right)$ is $n = 12$ and the number of terms in that particular expansion is $12+1 = 13.$ Since the problem asks for the coefficient of the middle term $C x^{18} y^{12},$ we need to find its middle term. Turns out, in the binomial expansion containing $13$ terms, the middle term would be the $7$th term.

Now looking for for the expression of the $7$th term:

$$nth \space term = C(n, r-1) u^{n-r+1} v^{r-1}$$ $$expression \space of \space 7th \space term = C(12, 7-1) (x^3)^{12-7+1} (2y^2)^{7-1}$$ $$ = (924) (x^3)^{6} (2y^2)^{6}$$ $$ = (924) (x^{18}) (2)^{6}(y^2)^{6}$$ $$ = (924) (x^{18}) (64) (y^2)^{6}$$ $$ = (924) (x^{18}) (64) (y^{12})$$ $$ = 59136 x^{18} y^{12}$$

Therefore, we conclude that $C = 59136.$

Lastly, the equivalent term of $(u^6)\left(\frac{v^6}{64} \right)$ from $(u + v)^n$, when we go back to dealing with $(x^3 + 2y^2)^n$, is $59136x^{18}y^{12}$

Hope it helps.....