Question

find the term containing x^10 in the expansion of (5+2x^2)^7

Answer 1

Answer:

x^10 = (x^2)^5

So you are looking at the term where the first term would be to the 5th power.

Since it is a 9th power binomial, you count down from 9

1st term ^9

2nd term ^8

3rd term ^7 ...

5th term ^5

So Pascals Triangle for a 9th power is

1 9 36 84 126 126 84 36 9 1

The 5th term has a coefficient of 126

It would be 126a^5b^4

Substituting from your binomial

126 * (2x^2)^5 * (-y^3)^4

=126*32x^10 * y^12

=4032 x^10 y^12

Step-by-step explanation:

I hope it's helping you