Question

A partial sum of an arithmetic sequence is given. find the sum.
1+5+9+...+401

Answer 1

ok, we have \(a_1=1\) and common difference is \(d=4\), last term is \(a_n=401\) and you sure know that nth term of the sequence (a_n) is given by:\[a_n=a_1+(n-1)d\]find \(n\)\[401=1+4(n-1) \\ n-1=100 \\ n=101\]and for sum use the formula\[S=\frac{n(a_1+a_n)}{2}=\frac{101(1+401)}{2}=101\times 201\]

Answer 2

What if it is a geometric sum?

Answer 3

For example: 1+3+9+...+2187
(thanks for the help on the other one by the way :) )

Answer 4

@mukushla :)

Answer 5

from that we get, a=, r=3, n = ?

Answer 6

Plugging some numbers here and there I get that
a_8=2187
now what?

Answer 7

nvm, I got it now xD it = 3280