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

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

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?