Evaluate the value of (1.02)^8 using binomial theorem up to two decimal places.

Step-by-step solution

Using the binomial theorem, (1 + x)^ n = 1 + nx + n(n-1) 2! x^2 + n(n-1)(n-2) 3! x^3 + Here, x = 0.02 and n = 8 . (1 + 0.02)^8 = 1 + 8(0.02) + 8 7 2 (0.02)^2 + 8 7 6 6 (0.02)^3 + Calculating the terms: First term: 1 Second term: 8 0.02 = 0.16 Third term: 28 0.0004 = 0.0112 Fourth term: 56 0.000008 = 0.000448 Summing these values: 1 + 0.16 + 0.0112 + 0.000448 = 1.171648 Rounding to two decimal places, we get 1.17 . Answer: 1.17