Question

I got this problem from an old Algebra book and the second part has got me beaten.
Part 1 What is the nth term of the sequence
1, 3/2, 5/4, 7/8 ......
Part 2
If the sum of n terms is denoted by s, write down the series obtained by subtracting 1/2 s from s, then find its value in terms of n. Hence find s.
I got part 1 to be (2n - 1) / 2^(n-1).

Answer 1

Specialise to find the pattern:

\(S_5 = 1 + \frac{3}{2} + \frac{5}{4}+ \frac{7}{8}+ \frac{9}{16}\)

\(\frac{S_5}{2} = ~~~~~\frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{ 16} + \frac{9}{32}\)

\(~~~\)

\(S_5 - \frac{S_5}{2} = 1 + (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} ) - \frac{9}{32}\)

Suggests
\(\frac{S_n}{2} = 1 + (\text{geometric series}) - \frac{2n-1}{2^{n}}\)

Geometric series starts at 1 and has 5 - 1 = 4 terms,

constant ratio \(\frac{1}{2}\)

Sum of geo series (n-1 terms) is:

\( \frac{1 - (\frac{1}{2})^{n-1}}{1 - \frac{1}{2}} = 2 - \frac{1}{2^{n-2}}\)


\(\therefore \frac{S_n}{2} = 1 + 2 - \frac{1}{2^{n-2}} - \frac{2n-1}{2^{n}}\)

\(= 3 - \frac{3 + 2n}{2^{n}} \)


\(\therefore S_n = 6 - \frac{3 + 2n}{2^{n-1}} \)

For the inductive proof:

\(S_{n+1} = 6 - \frac{3 + 2n}{2^{n-1}} \qquad + \frac{2(n+1)-1}{2^{n - 1 + 1}} \)

\( = 6 - \frac{6 + 4n}{2^{n}} + \frac{2n+1}{2^n}\)\)

\( = 6 - \frac{5 + 2n}{2^{n}} \)

\( = 6 - \frac{3 + 2(n+1)}{2^{n}} \)

Answer 2

Thank you very much. I got the first 3 lines then froze!!

Answer 3

Same!