KCSE 2025 Maths P2 Q4 — Binomial Expansion & Estimation

KCSE 2025 Form 4 Binomial Expansion

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The Question

“(a) Expand and simplify the expression (2 + 1 over 2x) all raised to the power 5. (b) Use the first four terms of the expansion in (a) to estimate the value of 2.05 raised to the power 5.”

1

Get the coefficients from Pascal's triangle

For a power of 5 the binomial coefficients come from the sixth row of Pascal's triangle. These numbers multiply each term as the power on the first part goes down from 5 to 0 and the power on the second part goes up from 0 to 5.

Row for power 5:15101051\text{Row for power } 5: \quad 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1
(2+12x)5=∑r=05(5r)25−r(12x)r\left(2 + \tfrac{1}{2x}\right)^{5} = \sum_{r=0}^{5}\binom{5}{r} 2^{5-r}\left(\tfrac{1}{2x}\right)^{r}
2

Write out every term

Take the two parts as 2 and 1 over 2x. Pair each Pascal coefficient with a falling power of 2 and a rising power of 1 over 2x. Writing all six terms in order sets up the expansion before any simplifying.

25+5⋅24(12x)+10⋅23(12x)2+10⋅22(12x)3+5⋅2(12x)4+(12x)52^{5} + 5\cdot 2^{4}\left(\tfrac{1}{2x}\right) + 10\cdot 2^{3}\left(\tfrac{1}{2x}\right)^{2} + 10\cdot 2^{2}\left(\tfrac{1}{2x}\right)^{3} + 5\cdot 2\left(\tfrac{1}{2x}\right)^{4} + \left(\tfrac{1}{2x}\right)^{5}
3

Simplify each term

Work out each power of 2 and each power of 1 over 2x, then multiply. The powers of 2 in the denominators (2, 4, 8, 16, 32) combine with the numbers in front, and the powers of x stay in the denominator. This gives the clean simplified expansion.

32+40x+20x2+5x3+58x4+132x532 + \frac{40}{x} + \frac{20}{x^{2}} + \frac{5}{x^{3}} + \frac{5}{8x^{4}} + \frac{1}{32x^{5}}
4

Choose x so the bracket equals 2.05

For part (b) the bracket 2 + 1 over 2x must equal 2.05. That forces 1 over 2x to be 0.05, so 2x is 20 and x is 10. This is the value that turns the whole expansion into 2.05 to the power 5.

2+12x=2.05⇒12x=0.052 + \frac{1}{2x} = 2.05 \Rightarrow \frac{1}{2x} = 0.05
2x=20⇒x=102x = 20 \Rightarrow x = 10
5

Substitute into the first four terms

Put x = 10 into just the first four terms of the expansion, as the question asks. Each term becomes a simple decimal, and adding them gives the estimate for 2.05 to the power 5.

32+4010+20102+510332 + \frac{40}{10} + \frac{20}{10^{2}} + \frac{5}{10^{3}}
=32+4+0.2+0.005=36.205= 32 + 4 + 0.2 + 0.005 = 36.205

Final Result

(a) The expansion is 32 + 40/x + 20/x^2 + 5/x^3 + 5/(8x^4) + 1/(32x^5). (b) Taking x = 10 in the first four terms gives 32 + 4 + 0.2 + 0.005 = 36.205, so 2.05 to the power 5 is approximately 36.205.

Why this method works

The binomial theorem works because expanding a bracket to the power 5 means choosing, from each of the five factors, either the first part or the second; Pascal's coefficients simply count how many ways each combination of powers can occur. Choosing x so the bracket equals 2.05 turns the algebraic identity into a numerical one, and because the later terms are tiny (they carry high powers of x in the denominator), the first four terms already give an accurate estimate.

✓

The true value 2.05^5 = 36.2073..., and our four-term estimate 36.205 agrees to three decimal places, confirming the working.