Odd-indexed terms and an alternating tail · source reference

Printed question

(a)(i) The sum of the first $n$ terms of a sequence $u_1,u_2,u_3,\ldots$ is $S_n=e^2n^2-3n$. Show $u_n=e^2(2n-1)-3$. [2]

(ii) Hence show that the sequence is an arithmetic progression. [2]

(iii) State the first 3 odd-numbered terms $u_1,u_3,u_5$. Hence find the least $n$ such that the sum of the first $n$ odd-numbered terms is more than 500. [3]

(b)(i) Find the range of $x$ for which the following geometric series converges: $e-e^{1+1/x}+e^{1+2/x}-e^{1+3/x}+\cdots$. [2]

(ii) Given $x=-1$, find the sum of all the terms of this geometric series after, and including, the 8th term. [3]

Printed working

(a)(i)

$$\begin{aligned}u_n&=S_n-S_{n-1}\\&=e^2n^2-3n-[e^2(n-1)^2-3(n-1)]\\&=e^2n^2-3n-[e^2(n^2-2n+1)-3n+3]\\&=e^2n^2-3n-e^2n^2+2e^2n-e^2+3n-3\\&=e^2(2n-1)-3.\end{aligned}$$

(ii) $u_n-u_{n-1}=e^2(2n-1)-3-[e^2(2n-3)-3]=2e^2$, constant, hence AP.

(iii) $u_1=e^2-3,u_3=5e^2-3,u_5=9e^2-3$. Common difference of odd-indexed terms $=4e^2$.

$$\frac n2[2(e^2-3)+(n-1)(4e^2)]>500.$$

From GC, $n=6$ gives 469.68 (<500), $n=7$ gives 651.4 (>500), so least $n=7$.

Alternatively sketch $y=\frac x2[2(e^2-3)+(x-1)4e^2]-500$. The positive $x$-intercept is approximately 6.18; the graph is positive for $n>6.18$. Least $n=7$.

(b)(i) Common ratio $r=\frac{-e^{1+1/x}}e=-e^{1/x}$.

$$|r|<1\Rightarrow|-e^{1/x}|<1\Rightarrow-1<e^{1/x}<1\Rightarrow e^{1/x}<1\Rightarrow\frac1x<\ln1=0\Rightarrow x<0.$$

(ii) At $x=-1$, $r=-e^{-1}=-1/e$.

Method 1:

$$S_{\infty}-S_7=\frac a{1-r}-\frac{a(1-r^7)}{1-r}=\frac{ar^7}{1-r}=\frac{e(-1/e)^7}{1-(-1/e)}=-0.00181\quad(3\text{ s.f.}).$$

Method 2: $u_8=ar^7=e(-1/e)^7=-1/e^6$.

$$\text{required sum}=\frac{u_8}{1-r}=\frac{-1/e^6}{1-(-1/e)}=-0.00181\quad(3\text{ s.f.}).$$

Teacher notes

$S_n=u_1+\cdots+u_n$, $S_{n-1}=u_1+\cdots+u_{n-1}$, so $u_n=S_n-S_{n-1}$. Replace $n$ by $n-1$ to obtain an expression for $u_n$.

To show AP, prove $u_n-u_{n-1}$ or $u_{n+1}-u_n$ is constant using general terms. Showing $u_3-u_2=u_2-u_1$ proves only those three terms are in AP.

If $u_1,u_2,u_3,\ldots$ has common difference $d$, the odd terms $u_1,u_3,u_5,\ldots$ form an AP with difference $2d$, since each spans two differences.

AP sum $S_n=\frac n2[2a+(n-1)d]$: here $a=e^2-3,d=4e^2$.

GC table: Y=, enter expression replacing $n$ by $x$; 2nd window → TblStart:1, ΔTbl:1; 2nd graph, scroll to rows changing from <500 to >500.

GC graph: Y=, replace $n$ by $x$; zoom → 0:ZoomFit. To find $x$-intercept: 2nd calc → 2:zero → left bound → right bound (move cursor on the curve) → zero.

Common ratio $r=u_2/u_1=u_3/u_2=\cdots$.

$S_{\infty}=u_1+u_2+\cdots$, $S_7=u_1+\cdots+u_7$. The sum after and including term 8 is $S_{\infty}-S_7=u_8+u_9+\cdots$.

Source tables and calculator displays

$n$	Odd-term sum
6	469.68 <500
7	651.4 >500

Source diagrams redrawn

Printed answer key

(a)(iii) $u_1=e^2-3,u_3=5e^2-3,u_5=9e^2-3$; 7. (b)(i) $x<0$; (ii) −0.00181 (3 s.f.)

Mathematical review and qualifications

(a)(i) $u_n=S_n-S_{n-1}=e^2(2n-1)-3$ (also correct at $n=1$). (ii) $u_{n+1}-u_n=2e^2$, constant. (iii) The odd-indexed terms begin $e^2-3,5e^2-3,9e^2-3$, with difference $4e^2$. Their sum is $n[(2n-1)e^2-3]$. At $n=6$ it is about $469.68$, at $n=7$ about $651.40$, so least $n=7$.

(b)(i) $|r|=e^{1/x}<1$ iff $1/x<0$, giving $x<0$. (ii) At $x=-1$, $r=-e^{-1}$, eighth term $er^7=-e^{-6}$. Its tail is $-e^{-6}/(1+e^{-1})\approx-0.00181$ (3 s.f.).

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2023/ACJC/Promo/8] Question PDF pages 1; solution PDF pages 4, 5, 6. The original files remain private QA records outside this native content.