What is the sum of the first 25 natural numbers?

Question

Part 2
if a_1 = 2 and r= 2, what are the first four terms of the geometric sequence?

Answer 1

To add the numbers 1 through n, proceed as follows. Pair up the first and last number to get a total of n+1. Then the second and second-to last number also sum to n+1. Proceed n/2 times if n is even, or (n-1)/2 times if n is odd; in the latter case, you have to add (n+1)/2 for the middle term. In either case, if you count how many factors of n+1 you have, you obtain the formula:
1 + 2 + 3 + ... + n = n(n+1)/2
Taking n = 25 gives 25*26/2 = 325.

There is a (possibly apocryphal) story that when Gauss was a child, his teacher needed some time to himself, so told all of his students to add up the numbers from 1 to 100. After a moment's reflection, Gauss devised the above strategy, and within a couple seconds had written down 101*50=5050 on his tablet, to the great surprise of his teacher.

Assuming you mean that the first term of your geometric sequence is a_1, and that the successive quotients are equal to r, the first four terms are:
a_1 = 2, a_2 = 4, a_3 = 8, a_4 = 16
If a_2/a_1 = 2 then a_2 = a_1*2 = 4, etc.

Answer 2

The formula for summing 1 through n is:

[n(n+1)] / 2.

So, if we let n = 25, the above formula evaluates as:

(25 x 26) / 2 = 650 / 2 = 325

Answer 3

if you have the numbers from 1 to n The sum is ((n+1)*n))/2 so if n = 25 S= 26*25/2 = 325

a_1=2 ,a_2=2 ,a_3=4 ,a_4=8

Answer 4

25x13=325


2 4 8 16

Answer 5

325!