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

Question

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

anonymous · Answer

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$$S=\frac{n(a_1+a_n)}{2}=\frac{101(1+401)}{2}=101\times 201$$

babynini · Answer

What if it is a geometric sum?

babynini · Answer

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

babynini · Answer

@mukushla :)

babynini · Answer

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

babynini · Answer

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

babynini · Answer

nvm, I got it now xD it = 3280