In a certain sequence, the next term is found by taking the number before it minus the number two numbers before it.
For example, in the sequence a, b, c, d... c = b-a, d = c-b, and so on.
Starting with 54 and 93, what would be the sum of the first six thousand terms?
VERSION 1 -
It's already been stated, but the sequence repeats itself every 6 terms, i.e.
1 - 54
2 - 93
3 - 93-54=39
4 - 39-93=-54
5 - -54-39=-93
6 - -93-(-54)=-39
7 - -39-(-93)=54
8 - 54-(-39)=93
Which brings us right back to where we started. The sum of the first 6 terms 54+93+39+(-54)+(-93)+(-39) = 0, thus since the sequence repeats every 6 terms the sum of any 6n terms, where n is a natural number is 0. For this problem let n=1000, so the sum of the first 6000 terms is 0.
VERSION 2 -
The sum of the first n terms-
n=1 sum=a
n=2 sum=a+b
n=3 sum= a+b+(b-a) = 2b
n=4 sum=2b+(c-b) = b+c
n=5 sum=b+c+(d-c) = b+d
n=6 sum=b+d+(e-d) = b+e
And so on. The 2nd term and the n-1 term remain after each successive addition. From part 1, the 2nd term is 93, and the n-1 term, with n=6000 is -93 (term 5). So the sum = 93+(-93) = 0.
2 ways to solve
