sum of first fifty?

Question

what is the sum of the first 50 positive integers

Answer 1

Sum = n/2 (t1 + t2) where n = total no. of terms, t1 = first term, t2 = last term

n=50
t1 = 1
t2= 50

sum = 1275

Answer 2

The sum of the first "n" consecutive positive integers can be found by: (1/2) * (n) * (n+1)

This is rather easily proven by induction, an outline of which will follow the answer to your specific question.

The sum of the first 50 positive integers is:
(1/2) * 50 * 51 = 1275.

Outline of proof: By induction.

Base case: n = 1. The sum of the first 1 positive integers is 1.

(1/2) * 1 * (1+1) = (1/2) * 2 = 1. So the base case works.

Now, assume that the sum of the first n positive integers is (1/2) * n * (n+1). Check if this holds for n+1:

The sum of the first n+1 integers is the sum of the first n integers plus (n + 1), and by our assumption, this would be

[(1/2) * n * (n+1)] + (n+1) = [n^2 + n + 2n + 2] / 2 = (1/2) * (n^2 + 3n + 2) = (1/2) * (n+1) * (n + 2).

Q.E.D.

Answer 3

1/2*50*51

Think of it this way:
Pair the 1 with the 50
Pair the 2 with the 49
Pair the 3 with the 48
Pair the 4 with the 47
.
.
.
Pair the 23 with the 28
Pair the 24 with the 27
Pair the 25 with the 26

Then you have 25 pairings each having a sum of 51.
So you have for an answer 25*51 or 1/2*50*51.

This works in general:
The sum of the first k integers is 1/2*k*(k+1).

Answer 4

The sum of the first n positive integers is n(n+1)/2.
So the sum of the first 50 is 50*51/2 = 1275.
How do we get the formula for the sum of the
first n positive integers?
Write
1 + n = n+1
2 + (n-1) = n+1
3 + (n-2) = n+1
--------------------
n + 1 = n+1
Call the sum we want to find S.
Adding all the columns, we find
S + S = n(n+1)
So
S = n(n+1)/2.

Answer 5

Gauss solved that when he was a kid.

Look at the number series:

1, 2, 3, ..., 50

Look at it backwards:

50, 49, 48, ...1

Add the two together:

(1+50), (2+49), (3+48), ...(50+1)

Note that each term in this new series equals 51, and there are 50 terms in the series. So the total of all the terms is 51*50. But you used each number twice, so divide it by two. That is, the sum of the digits 1 through 50 equals 51*50/2.

Answer 6

aside from 50...
take

1+49 =50
2+48 =50
3+47 =50
.
.
.
.
24+26 =50
theres 24 of these combinations



25 will be left free standing with 50

hence: (24*50) + 25 + 50 = 1275

Answer 7

I wrote a small script to come up with this answer, I had it add all the numbers from 0 to 50 (including 50) and it came back saying: 1275. If you save the following code in an HTML file you'll see the results I recieved:





(I ran the script in Internet Explorer 7.)

Answer 8

Sn = (n/2)(a1 + an)

S(50) = (50/2)(1 + 50)
S(50) = 25(51)
S(50) = 1275

Answer 9

25*50+25? ~

Answer 10

Use: n(n+1)/2
(the highest number)(the highest number plus one) then divide by two

n=50 (here)

50(50+1)=2550
2,550/2= 277

Total= 277