# Ap/GP Question

1. Apr 8, 2005

### al_201314

Hi everyone,

This is my first post here Anyway I have problems solving this question wonder anyone could help give me some clues as to how to go about it. Here goes:

The positive integers are bracketed as follows,

(1), (2,3), (4,5,6,7), (8,9,10,11,12,13,14,15), ...

No. of integers in the rth bracket is 2^r-1. State stae first term and last term in the nth bracket. Hence, show that the sum of the terms in the first n brackets is 2^n-1[(2^n)]-1].

Thanks very much.. I have absolutely no clue..

alvin

2. Apr 8, 2005

### StatusX

Start by finding how many numbers are in the first n brackets, which is just the sum of the first n powers of 2 (use the geometric sum formula). This will let you know the first and last number in each bracket. Then, to find the sum, use the sum of the first n positive integers, and plug in what you just found for n. If you don't know the sum of the first n integers, the easiest way to find it is to use a telescoping sum:

$$\sum_{k=1}^{n} (k+1)^2 - k^2$$

If you were to write this out term by term, you'd see each term cancels except the first and last. Now that you have this sum, rewrite it as:

$$\sum_{k=1}^{n} k^2+ 2k + 1 - k^2 = \sum_{k=1}^{n} 2k + 1$$

from which the result should follow pretty easily.

Last edited: Apr 8, 2005
3. Apr 9, 2005

### al_201314

I got it.. thanks status.