Apply the binomial theorem.

Here's the solution to your question: a) i) Write down the binomial expansion of (x+2y)^5. Step 1: Apply the binomial theorem. The binomial theorem states that (a+b)^n = _k=0^n nk a^n-k b^k. For (x+2y)^5, we have a=x, b=2y, and n=5. The binomial coefficients for n=5 are 50=1, 51=5, 52=10, 53=10, 54=5, 55=1. Step 2: Expand the expression. (x+2y)^5 = 50x^5(2y)^0 + 51x^4(2y)^1 + 52x^3(2y)^2 + 53x^2(2y)^3 + 54x^1(2y)^4 + 55x^0(2y)^5 (x+2y)^5 = 1 · x^5 · 1 + 5 · x^4 · (2y) + 10 · x^3 · (4y^2) + 10 · x^2 · (8y^3) + 5 · x · (16y^4) + 1 · 1 · (32y^5) (x+2y)^5 = x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5 The binomial expansion is x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5. ii) Use your expansion in (i) to evaluate (1.02)^5, correct to three decimal places. Step 3: Relate (1.02)^5 to (x+2y)^5. We can write 1.02 as 1 + 0.02. Comparing (x+2y)^5 with (1+0.02)^5, we can set x=1 and 2y=0.02. From 2y=0.02, we get y=0.01. Step 4: Substitute x=1 and y=0.01 into the expansion. (1.02)^5 = (1)^5 + 10(1)^4(0.01) + 40(1)^3(0.01)^2 + 80(1)^2(0.01)^3 + 80(1)(0.01)^4 + 32(0.01)^5 Calculate each term: (1)^5 = 1 10(1)^4(0.01) = 10 × 1 × 0.01 = 0.1 40(1)^3(0.01)^2 = 40 × 1 × 0.0001 = 0.004 80(1)^2(0.01)^3 = 80 × 1 × 0.000001 = 0.00008 80(1)(0.01)^4 = 80 × 0.00000001 = 0.0000008 32(0.01)^5 = 32 × 0.0000000001 = 0.0000000032 Step 5: Sum the terms and round to three decimal places. (1.02)^5 = 1 + 0.1 + 0.004 + 0.00008 + 0.0000008 + 0.0000000032 (1.02)^5 = 1.1040808032 Rounding to three decimal places, we look at the fourth decimal place. Since it is 0 (which is less than 5), we keep the third decimal place as it is. (1.02)^5 ≈ 1.104 The value of (1.02)^5 correct to three decimal places is 1.104. 3 done, 2 left today. You're making progress.