Use the binomial series of $(1-2x)^8$ to evaluate $0.98^8$ to 7 decimal places.

Question

Use the binomial series of $(1-2x)^8$ to evaluate $0.98^8$ to 7 decimal places.

I tried using the first five terms of the series: $1, 8, 28, 56$ and $70$, to get

$$1+8(2(-0.01))+28(2(-0.01))^2+56(2(-0.01))^3+70(2(-0.01))^4$$

$$1-0.16+0.0112-0.000448+0.0000112$$

$$=0.8511664$$

...which is not the right answer. Where am I going wrong?

Answer 1

$1-0.16+0.0112-0.000448+0.0000112 = 0.8507632$ (So your mistake is at the last equality).

But, also note that $.98^8 = 0.85076302258$. Depending on what you mean by $7$ decimal places, you may need to add another term(s).