Odd-indexed terms and an alternating tail

(a)(i) The sum of the first nn terms of a sequence u1,u2,u3,…u_1,u_2,u_3,\ldots is Sn=e2n2−3nS_n=e^2n^2-3n. Show un=e2(2n−1)−3u_n=e^2(2n-1)-3. marks

(ii) Hence show that the sequence is an arithmetic progression. marks

(iii) State the first 3 odd-numbered terms u1,u3,u5u_1,u_3,u_5. Hence find the least nn such that the sum of the first nn odd-numbered terms is more than 500. marks

(b)(i) Find the range of xx for which the following geometric series converges: e−e1+1/x+e1+2/x−e1+3/x+⋯e-e^{1+1/x}+e^{1+2/x}-e^{1+3/x}+\cdots. marks

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

[2023/ACJC/Promo/8]

Total marks: 12

Solution

(a)(i)

Terms from sums

Sn=e2n2−3nS_n=e^2n^2-3n. Show un=e2(2n−1)−3u_n=e^2(2n-1)-3.

Hint

  1. un=Sn−Sn−1u_n=S_n-S_{n-1}.

Answer · (a)(i)

un=e2(2n−1)−3u_n=e^2(2n-1)-3.

(a)(ii)

Arithmetic progression

Show the sequence is an AP.

Hint

  1. Show un−un−1u_n-u_{n-1} is constant.

Answer · (a)(ii)

AP with difference 2e22e^2.

(a)(iii)

Odd-numbered terms

State u1,u3,u5u_1,u_3,u_5 and find the least nn such that the sum of the first nn odd-numbered terms exceeds 500.

Explore this idea

Hint

  1. The odd-numbered terms form an AP with difference 4e24e^2.

  2. Solve n2[2(e2−3)+(n−1)4e2]>500\frac n2[2(e^2-3)+(n-1)4e^2]>500.

Answer · (a)(iii)

u1=e2−3,u3=5e2−3,u5=9e2−3u_1=e^2-3,u_3=5e^2-3,u_5=9e^2-3; least n=7n=7.

Explore the idea

The sum of the first n odd-numbered terms

Drag the slider to change n and compare the plotted values.

Indexn = 7
Try an example
sum of n odd terms at n = 7
651.4041
Key idea

The sum passes 500 between n=6n=6 and n=7n=7: the least nn is 7.

Ideas to try
Find the first nn above 500.
(b)(i)

Convergence of a GP

Find the range of xx for which e−e1+1/x+e1+2/x−⋯e-e^{1+1/x}+e^{1+2/x}-\cdots converges.

Hint

  1. r=−e1+1/xe=−e1/xr=\frac{-e^{1+1/x}}{e}=-e^{1/x}.

Answer · (b)(i)

x<0x<0.

(b)(ii)

Tail of a GP

With x=−1x=-1, find the sum of all the terms from the 8th onwards.

Explore this idea

Hint

  1. r=−1er=-\frac1e, a=ea=e.

  2. Tail sum =u81−r=\frac{u_8}{1-r}.

Answer · (b)(ii)

−e−6/(1+e−1)≈−0.00181-e^{-6}/(1+e^{-1})\approx-0.00181.

Explore the idea

Terms of the alternating GP (x = −1)

Drag the slider to change n and compare the plotted values.

Indexn = 7
Try an example
term uₙ at n = 7
0.006738
sum Sₙ at n = 7
1.989
Key idea

The running total oscillates above and below its limit e1+1/e≈1.99\frac e{1+1/e}\approx1.99, getting closer each time.

Ideas to try
The terms alternate in sign and shrink: how does that affect the running total?
Takeaway

“Including term 8” means subtract seven terms

A tail beginning at index kk equals S∞−Sk−1=uk/(1−r)S_\infty-S_{k-1}=u_k/(1-r). An odd-indexed AP subsequence has twice the original difference.