Question: Find the value of (0.98)^1/4 correct to 5 decimal places.
Use the binomial expansion.
(1 + x)^n = 1 + nx + n(n-1)/2* + n(n-1)(n-2)/3!*x^3 + ...
So for (1 - 0.02)^(1/4), the expansion is,
(1 + (-0.02))^(1/4) = 1 + (1/4)*(-0.02) + (1/4)(1/4 - 1)/2*(-0.02)^2 + (1/4)(1/4 - 1)(1/4 - 2)/3!*(-0.02)^3 + ...
(1 + (-0.02))^(1/4) = 1 - 0.005 - 0.0000375 - 0.00000004375
(0.98)^(1/4) = 0.9949620625
(0.98)^(1/4) = 0.99496