Question
2+4+8+...is a geometric sequence. Prove the sum of the first n terms is two less than the (n+1)the term.
Answers
Reiny
a = 2 , r = 2
sum(n) = 2(2^n - 1)/(2-1) = 2(2^n - 1)
= 2^(n+1) - 2
sum(n+1) = 2(2^(n+1) - 1)/1 = 2(2^(n+1) - 1)
= 2^(n+2) - 2
sum(N+1) - sum(n) =2^(n+2) - 2 - (2^(n+1) - 2)
= 2^(n+2) - 2^(n+1)
= 2^(n+1) (2 - 1)
= 2^(n+1) , which is a positive number
so sum(n) < sum(n+1)
sum(n) = 2(2^n - 1)/(2-1) = 2(2^n - 1)
= 2^(n+1) - 2
sum(n+1) = 2(2^(n+1) - 1)/1 = 2(2^(n+1) - 1)
= 2^(n+2) - 2
sum(N+1) - sum(n) =2^(n+2) - 2 - (2^(n+1) - 2)
= 2^(n+2) - 2^(n+1)
= 2^(n+1) (2 - 1)
= 2^(n+1) , which is a positive number
so sum(n) < sum(n+1)