Final answer:
The binomial expansion of (1.05)⁵ is 2.0887 when rounded to four decimal places.
Step-by-step explanation:
To find the binomial expansion of (1.05)⁵, we can use the binomial theorem formula:
(a + b)ⁿ = C(n, 0)aⁿb⁰ + C(n, 1)aⁿ⁻¹b + C(n, 2)aⁿ⁻²b² + ... + C(n, n-1)abⁿ⁻¹ + C(n, n)a⁰bⁿ
In this case, a = 1.05, b = 0.05, and n = 5. Plugging these values into the formula, we get:
(1.05 + 0.05)⁵ = C(5, 0)(1.05)⁵(0.05)⁰ + C(5, 1)(1.05)⁴(0.05)¹ + C(5, 2)(1.05)³(0.05)² + C(5, 3)(1.05)²(0.05)³ + C(5, 4)(1.05)¹(0.05)⁴ + C(5, 5)(1.05)⁰(0.05)⁵
Simplifying further:
(1.05 + 0.05)⁵ = (1)(1.05)⁵(0.05)⁰ + (5)(1.05)⁴(0.05)¹ + (10)(1.05)³(0.05)² + (10)(1.05)²(0.05)³ + (5)(1.05)¹(0.05)⁴ + (1)(1.05)⁰(0.05)⁵
Calculating the values:
(1)(1.05)⁵(1) + (5)(1.05)⁴(0.05) + (10)(1.05)³(0.05)² + (10)(1.05)²(0.05)³ + (5)(1.05)¹(0.05)⁴ + (1)(1)(0.05)⁵
Rounding each term to four decimal places:
1.2763 + 0.6401 + 0.1503 + 0.0210 + 0.0010 + 0.0000
Adding up all the terms:
1.2763 + 0.6401 + 0.1503 + 0.0210 + 0.0010 + 0.0000 = 2.0887
Finally rounding the final answer to four decimal places, we get 2.0887 rounded to 2.0887.