For the answer to the question above,
1 + nx + [n(n-1)/(2-factorial)](x)^2 + [n(n-1)(n-2)/3-factorial] (x)^3
1 + nx + [n(n-1)/(2 x 1)](x)^2 + [n(n-1)(n-2)/3 x 2 x 1] (x)^3
1 + nx + [n(n-1)/2](x)^2 + [n(n-1)(n-2)/6] (x)^3
1 + 9x + 36x^2 + 84x^3
In my experience, up to the x^3 is often adequate to approximate a route.
(1+x) = 0.98
x = 0.98 - 1 = -0.02
Substituting:
1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3
approximation = 0.834
Checking the real value in your calculator:
(0.98)^9 = 0.834
So you have approximated correctly.
If you want to know how accurate your approximation is, write out the result of each in full:
1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 = 0.833728
(0.98)^9 = 0.8337477621
So it is correct to 4
