A.What is the fourth term of (3/2y^2+2/3y^-3)^7?
B. Consider the binomial product (a+b)^n where the third term of the expansion is 127575x^5y^4 and the fourth term of the expansion is -354375x^4y^6. Find a,b, and n.

Question

A.What is the fourth term of (3/2y^2+2/3y^-3)^7?
B. Consider the binomial product (a+b)^n where the third term of the expansion is 127575x^5y^4 and the fourth term of the expansion is -354375x^4y^6. Find a,b, and n.

campbell_st · Answer

just to make life easy is the binomial 

$$(\frac{3}{2} y^2 + \frac{2}{3} y^{-3})^7$$

anonymous · Answer

Yes that is binomial (:

campbell_st · Answer

ok... so the 4th term is 

$$^7C_{3} (\frac{3}{2}y^2)^{7 -3} \times (\frac{2}{3} y^{-3})^3$$

which becomes 

$$35 \times \frac{3^4}{2^4}(y^2)^4 \times \frac{2^3}{3^3}(y^{-3})^3$$

you need to simplify it... 

as far as the fractions go you'll have 

$$\frac{3^4}{2^4} \times \frac{2^3}{3^3} = \frac{3}{2}$$

hope it helps

anonymous · Answer

Thank you! This does help. How did you get the 35? and is the y with the negative exponent to a power of 3 or two?

campbell_st · Answer

well 35 is the 7C3 combination..... 

you can also find the coefficient of 35 from pascals triangle

anonymous · Answer

perfect. thank you!