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IAL 2023 Oct Q10

A Level / Edexcel / P2

IAL 2023 Oct Paper · Question 10

题目

Problem

(i) (a) Find, in ascending powers of xx, the 2nd, 3rd and 5th terms of the binomial expansion of (3+2x)6(3 + 2x)^6.

For a particular value of xx, these three terms form consecutive terms in a geometric series.

(b) Find this value of xx.

(ii) In a different geometric series,

  • the first term is sin⁡2θ\sin^2\theta
  • the common ratio is 2cos⁡θ2\cos\theta
  • the sum to infinity is 85\dfrac{8}{5}

(a) Show that 5cos⁡2θ−16cos⁡θ+3=05\cos^2\theta - 16\cos\theta + 3 = 0.

(b) Hence find the exact value of the 2nd term in the series.

(12)
题目中文翻译

(i) (a) 求 (3+2x)6(3 + 2x)^6 的二项式展开式中，按 xx 的升幂排列的第2项、第3项和第5项。

对于某个特定的 xx 值，这三项构成等比数列的连续三项。

(b) 求这个 xx 值。

(ii) 在另一个等比数列中，

  • 首项为 sin⁡2θ\sin^2\theta
  • 公比为 2cos⁡θ2\cos\theta
  • 无穷项和为 85\dfrac{8}{5}

(a) 证明 5cos⁡2θ−16cos⁡θ+3=05\cos^2\theta - 16\cos\theta + 3 = 0。

(b) 由此求该数列第2项的精确值。

（12分）

解答

解法一

思路

(i)(a) 利用二项式定理展开，第 r+1r+1 项为 (nr)an−rbr\binom{n}{r} a^{n-r} b^r。 (b) 连续三项成等比数列意味着 b2=acb^2 = ac。 (ii)(a) 无穷等比数列求和公式 S∞=a1−rS_\infty = \dfrac{a}{1 - r}。 (b) 第二项为 arar。

答题过程

(i)(a) The general term of (3+2x)6(3 + 2x)^6 is Tr+1=(6r)36−r(2x)rT_{r+1} = \binom{6}{r} 3^{6-r} (2x)^r.

T_2 =&\, \binom{6}{1} 3^5 (2x) = 6 \times 243 \times 2x = 2916x \\[2mm] T_3 =&\, \binom{6}{2} 3^4 (2x)^2 = 15 \times 81 \times 4x^2 = 4860x^2 \\[2mm] T_5 =&\, \binom{6}{4} 3^2 (2x)^4 = 15 \times 9 \times 16x^4 = 2160x^4 \end{align*}$$ $$\boxed{T_2 = 2916x, \quad T_3 = 4860x^2, \quad T_5 = 2160x^4}$$ **(i)(b)** Since these three terms are consecutive terms of a geometric series: $$\begin{align*} T_3^2 =&\, T_2 \times T_5 \\ (4860x^2)^2 =&\, 2916x \times 2160x^4 \\ 23619600x^4 =&\, 6298560x^5 \\ x =&\, \frac{23619600}{6298560} = \frac{15}{4} \end{align*}$$ $$\boxed{x = \frac{15}{4}}$$ **(ii)(a)** For the geometric series with first term $a = \sin^2\theta$, common ratio $r = 2\cos\theta$, and $S_\infty = \dfrac{8}{5}$: $$\begin{align*} \frac{\sin^2\theta}{1 - 2\cos\theta} =&\, \frac{8}{5} \\[2mm] 5\sin^2\theta =&\, 8(1 - 2\cos\theta) \\ 5(1 - \cos^2\theta) =&\, 8 - 16\cos\theta \\ 5 - 5\cos^2\theta =&\, 8 - 16\cos\theta \\ 5\cos^2\theta - 16\cos\theta + 3 =&\, 0 \end{align*}$$ $$\boxed{5\cos^2\theta - 16\cos\theta + 3 = 0}$$ **(ii)(b)** Solving $5\cos^2\theta - 16\cos\theta + 3 = 0$: $$\begin{align*} \cos\theta =&\, \frac{16 \pm \sqrt{256 - 60}}{10} = \frac{16 \pm 14}{10} \end{align*}$$ $\cos\theta = 3$ is rejected (since $|\cos\theta| \leqslant 1$), so $\cos\theta = \dfrac{1}{5}$. The second term of the series is $ar = \sin^2\theta \times 2\cos\theta$: $$\begin{align*} \sin^2\theta =&\, 1 - \cos^2\theta = 1 - \frac{1}{25} = \frac{24}{25} \\[2mm] T_2 =&\, \frac{24}{25} \times 2 \times \frac{1}{5} = \frac{48}{125} \end{align*}$$ $$\boxed{T_2 = \frac{48}{125}}$$