If you add up the number 1 - 100 consecutively yourr total is?

Answer 1

4528, I am pretty sure. I just did it on my calculater but I could have messed up. Nevermind, I did it again and it is 5050.

Answer 2

5050

1 + 100 is 101
2 + 99 is 101
3 + 98 is 101

Since there are 100 numbers (or 50 combinations) which total 101 as you proceed inward from 1 to 100, just multiply 101 by 50 and you come out with the answer 5050.

Answer 3

The sum is 5050. Why?:

1, 2, 3, ... 99, 100 is an ARITHMETIC SERIES. As such, its sum is given by :

S = (1/2) n (2 a + [n - 1] d), *** where ' n ' is the no. of terms (100), ' a ' is the first term (1 here), and ' d ' is the (common) difference between successive terms (1 again, in this case). Hence:

S = 50 (2 + 99) = 50 (101) = 5050.

(This one was particularly easy; it could be done in one's head.)

Live long and prosper.

*** POSTCRIPTS

1. The general formula is proved by writing "S = ... + ... + ... " (the sum of the separate generally written terms in the series) on a first line, ### summing from left to right (that is, from ' a ' to ' a + [n - 1] d '), and then repeating that again underneath ("S = ... + ...+ ... "0, in reversed order (summing now from ' a + [n - 1]d ' to ' a '). Adding the TWO series now gives the sum 2 S on the LHS and ' n ' separately added terms, all of ' a + (a + [n - 1]d ' on the RHS. So:

2 S = n (2a + [n - 1] d), or:

S = (1/2) n (2a + [n - 1] d) ........ QED

Students who have studied series are expected to know this result by heart. (I still do, 57 years later.)

### It is difficult to do this horizontal display explicitly in these Answers, because of Yahoo's font spacing limitations. A solution to this problem is to write the GENERAL series VERTICALLY, as 'answeringyou' did explicitly for this particular series sum. The general expression for the sum is derived analogously to her explicit solution. Indeed, she probably used that model for her demonstration.

2. ' Answerer ' reminds me that young Gauss's teacher thought that he would keep his class busy for a whole class period, adding up these numbers, so that he could get on with some marking, undisturbed. But Gauss's hand shot up very quickly with the answer. He had in effect discovered the general rule. He excelled --- WITHOUT Excel!

Answer 4

5050. This is a famous solution solved by Gauss when he was in elementary school. You have to notice that:

1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
..

101 x 50 = 5050

Answer 5

When you have got numbers listed consecutively, then the simplest way to find the sum is:

n * (n + 1) / 2

So, when n = 100, the sum would be:

100 * 101/2 = 5050

You can modify this method to find the sum of all numbers from 31 to 50!! Try it and let us know what you get!!

Good luck.

Answer 6

Sum from 1 to n of x = n(n+1)/2
so sum from 1 to 100 is 100*101/2 = 50*101=5050

Answer 7

d o you mean 1+1+2+3+4+5+6+7+8+9+10.....100?
thats easy you can do it with any number
S = 1/2 N ((n+n) + (N-n) ..............here n=1 and N=100
S= 50 (2 + 99) = 5050

Answer 8

5050
Just break it up, like this:
1+100=101
2+99=101
3+98=101
etc...
there are 50 pairs of these so it is 101 times 50, and there is your answer!

Answer 9

your total is 5,050 consequently

Answer 10

there is a formula... (i think.)

if n is the number of terms (numbers) and d is the difference between each one (1 in this case) then
total=(first term + last term)/half n

so thats 5050