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
Jhun Vert

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

- 983 reads

Subscribe to MATHalino on

- Tapered Beam
- Vickers hardness: Distance between indentations
- Time rates
- Minima Maxima: y=ax³+bx²+cx+d
- Make the curve y=ax³+bx²+cx+d have a critical point at (0,-2) and also be a tangent to the line 3x+y+3=0 at (-1,0).
- Minima maxima: Arbitrary constants for a cubic
- Minima Maxima: 9a³y=x(4a-x)³
- Minima maxima: a²y = x⁴
- how to find the distance when calculating moment of force
- strength of materials
- Analytic Geometry Problem Set [Locked: Multiple Questions]
- Equation of circle tangent to two lines and passing through a point
- Product of Areas of Three Dissimilar Right Triangles
- Perimeter of Right Triangle by Tangents
- Differential equations
- Laplace
- Families of Curves: family of circles with center on the line y= -x and passing through the origin
- Family of Plane Curves
- Differential equation
- Differential equation

Home • Forums • Blogs • Glossary • Recent

About • Contact us • Disclaimer • Privacy Policy • Hosted by WebFaction • Powered by Drupal

About • Contact us • Disclaimer • Privacy Policy • Hosted by WebFaction • Powered by Drupal

Forum posts (unless otherwise specified) licensed under a Creative Commons Licence.

All trademarks and copyrights on this page are owned by their respective owners. Forum posts are owned by the individual posters.

All trademarks and copyrights on this page are owned by their respective owners. Forum posts are owned by the individual posters.

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.....