# 1-100 sum???????????

adding all the numbers 1-100 together, what would be the answer? How do you figure other then just adding them together individually.

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• Anonymous

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

=

1+ 100 + (this is the 1st row)

2 + 99 +

3 + 98 +

4 + 97 +

...

50 + 51 (this is the 50th row)

=

101 + (this is the 1st row)

101 +

..... + (rows 3 to 49 go here, all say "101"...)

101 (this is the 50th row)

= 101 * 50 =

Another way. Let

(A) 1 + 2 + 3 + .... 98 + 99 + 100 = S

(there are 100 numbers being added here)

(B) 100 + 99 + 98 + ... + 3 + 2 + 1 = S

(there are again 100 numbers being added here),

Then add the first term of (A) to the first term of (B). Add the second term of (A) to the second term of (B). Continue until you get to the 100th term. Notice that each sum is 101. There sre 100 such sums.

101 + 101 + ... 101 (the 100th one) = 2S.

101 + 101 + ... + 101 (the 100th one) = 101*100 = 10100 = 2S

Therefore S = 10100/2 = 5050 = ANSWER

Gauss supposedly figured that out when he was like 5 years old. I don't think anyone else could have! This is a hard question.

Finally, for future reference:

1 + 2 + 3 + ... + n = n(n+1)/2. This works for even or odd n.

Hope that helped!!! Good luck on your hw. =)

• Sum Of 1 To 100

• Well, you write down the first few numbers and the last few numbers. I take the first three as an example, but you might want to take four or five.

1, 2, 3, ..., 98, 99, 100

You look at this as long as it takes until you suddenly see (your mind tells you) that the first and the last add up to 101, and the second and the next-to-last add up to 101, and the third and the next-to-next-to-last add up to 101.

Then you ask yourself (out of natural curiosity) whether this could be continued. And you figure out that if you add 1 to the first number in a sum and take away 1 from the second number in the sum, then the sum cannot change.

You realize that you only have to calculate how many pairs like 1+100, 2+99, and 3+98 (all equal to 101) will be formed from the numbers 1 to 100. Oh, you say, "One hundred numbers arrange into fifty pairs!"

So the total sum of the number from 1 to 100 must 50 times 101, which is exactly 5050.

1) there is a formula for sums: n(n+1)/2

so 100(101)/2 =10,100/2 = 5,050

2) imagine adding 1 + 2 + 3 + ...+100 = N

then think of 100 + 99 + 98 + ..... + 1 = N

of course N will be the same either way.

adding together both sums gives you

(1 + 100) + (2 + 99) + (3 + 98) + ...+(100 + 1) = 2N

but notice that each amount in parentheses equals 101,

and that there are 100 groups in parentheses,

so the left side equals (100)(101) = 2N

10,100 = 2N

5,050 = N

this is the reasoning that leads to the first formula I gave