Expand (1+2x)^20 - (1-2x)^20) in ascending powers of x up to the term in x^5 and use this result to evaluate (1.02)^20 - (0.98)^20 to 3 significant figures.

Answer 1

(1+2x)^20 - (1-2x)^20)

=(1 + 20(2x) + 190(4x^2 + 1140(8x^3) + 4845(16x^4) + 15504(32x^5 + ...) - (1 - 20(2x) + 190(4x^2) - 1140(8x^3) + 4845(16x^4) - 15504(32x^5) + ..)
= 80x + 18240x^3 + 992256x^5

comparing (1.02)^20 - (0.98)^20 with (1+2x)^20 - (1-2x)^20
we can see that x = .01

so if x = .01
80x + 18240x^3 + 992256x^5
= 80(.01) + 18240(.00001) + 992256(.000000001)
= .8 +01824 + .0000992256
= .818339225
= .818 using 3 significant figures

real answer: .818339425

As I said before, a rather futile exercise, merely showing the mathematical property but not at all practical.
If x had not been such a nice number like .01, but rather something like x = .31 we would still have a horrible calculation to do

Answer 2

In the expansion of (1-2x)^11 the coefficient of x^3 is k times the coefficient of x^2. Evaluate k.