Odd-sized groups of geometric terms

The sum of the first nn terms, SnS_n, is Sn=1a[1−(a−1)n]S_n=\frac1a[1-(a-1)^n], where aa is a constant, a≠1a\ne1, n∈Z+n\in\mathbb Z^+. Obtain the nnth term TnT_n and prove that “SnS_n is a geometric series” (printed wording). marks

The sequence {Tn}\{T_n\} is grouped as (T1),(T2,T3,T4),(T5,T6,T7,T8,T9),…(T_1),(T_2,T_3,T_4),(T_5,T_6,T_7,T_8,T_9),\ldots, each subsequent bracket having 2 more terms than the previous. Find:

(i) total terms in the first nn brackets marks;

(ii) the middle term of bracket 11 in terms of aa marks;

(iii) the range of aa for the sum to infinity to exist; and the least nn for the sum of all terms in the first nn brackets to be within 0.1% of the sum to infinity when a=3920a=\frac{39}{20}. marks

[modified 2011/AJC/Prelim/I/12]

Total marks: 11

Solution

Terms from sums

Recover the GP terms

Obtain TnT_n from the given SnS_n and identify the geometric progression.

Hint

  1. Subtract the two powers after factoring out (a−1)n−1(a-1)^{n-1}.

Answer · Terms from sums

Tn=(2−a)(a−1)n−1/aT_n=(2-a)(a-1)^{n-1}/a, GP ratio a−1a-1.

a≠0,1a\ne0,1.

(i)

Count grouped terms

Count the terms in the first nn brackets.

Hint

  1. The nth bracket has 2n−12n-1 terms.

Answer · (i)

n2n^2 terms.

(ii)

Locate the middle term

Find the middle term of bracket 11.

Hint

  1. Use 100+11100+11, not the bracket number as the GP index.

Answer · (ii)

T111=(2−a)(a−1)110/aT_{111}=(2-a)(a-1)^{110}/a.

(iii)

Convergence and grouped error

Find the convergence range, then the least number of brackets giving relative error below 0.1%0.1\% when a=39/20a=39/20.

Explore this idea

Hint

  1. The required relative error is 0.1%=0.001.

  2. How many terms are actually included in n brackets?

Answer · (iii)

Literal convergence: 0<a≤20<a\le2, a≠1a\ne1; least grouped count n=12n=12.

The printed 0<a<20<a<2 omits the zero series at a=2a=2; a=0a=0 is undefined.

Explore the idea

A bracket count changes the exponent

At a = 39/20, compare the remaining fraction after n brackets, which contain n² terms.

Bracket countn = 12
Try an example
relative tail at n = 12
0.0006197
Key idea

Eleven brackets leave about 0.202%; twelve leave about 0.062%.

Ideas to try
Compare the values on either side of the boundary. What makes one count qualify?
Takeaway

Count terms before applying a GP formula

The partial sum of n brackets is Sn2S_{n^2}; substituting n instead would miss most terms.