Binomial expansion problem

[iitsrii]

First 3 terms of (4-2x^-2)^5
first term is 4^5 which is 1024
4^4(1/2x^2) = 256/2x^-2
128/x^2
3rd term
64(1/2x^2)^2 = 64/2x^2
32/x
not sure about the 3rd term

 

[tkhunny]

First 3 terms of (4-2x^-2)^5
first term is 4^5 which is 1024
4^4(1/2x^2) = 256/2x^-2
128/x^2
3rd term
64(1/2x^2)^2 = 64/2x^2
32/x
not sure about the 3rd term

In a pinch, with only a few terms, just write them ALL down. This will also help you internalize the formula for immediately writing a single term.

Binomial coefficients
1
5
10
10
5
1

Powers of First Term (Descending Exponents)
1 * 4^5
5 * 4^4
10 * 4^3
10 * 4^2
5 * 4^1
1 * 4^0

Powers of Second Term (Ascending Exponents)
1 * 4^5 * (-2/(x^2))^0
5 * 4^4 * (-2/(x^2))^1
10 * 4^3 * (-2/(x^2))^2
10 * 4^2 * (-2/(x^2))^3
5 * 4^1 * (-2/(x^2))^4
1 * 4^0 * (-2/(x^2))^5

Pretty close to done with anything that could be asked.