1 + 2 + 3 +..... +100 = ?

Question

equals?

Answer 1

100......

Answer 2

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
+ 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 = 145
+ 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 = 245
+ 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 = 345
+ 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 = 445
+ 50 + 51 + 52 + 53 + 54 + 55 + 56 + 57 + 58 + 59 = 545
+ 60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 = 645
+ 70 + 71 + 72 + 73 + 74 + 75 + 76 + 77 + 78 + 79 = 745
+ 80 + 81 + 82 + 83 + 84 + 85 + 86 + 87 + 88 + 89 = 845
+ 90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 = 945
+ 100 = 5050

Answer 3

5050

You can see how this is true by adding pairs of numbers from the list. The first (1) and the last (100) add up to 101. The second (2) and the second-to-last (99) add up to 101, and so on.

Since the list is 100 numbers long, you have half that many pairs of numbers - 50 pairs.

To add 50 pairs of 101 each - 50 x 101 = 5050.

This trick was first worked out by a famous German mathematician named Gauss - when he was less than 10 years old.

In general, the sum of any string of numbers 1 + 2 + 3 + ... + N is equal to:

[N x (N+1)] / 2

Answer 4

When Gauss was at primar school, the math teacher asked the children the above to calculate. 1+2+3+...........+100=?
All the other students were calculating adding one after another. Gauss almost immediately gave the answer 5050. The teacher was suprised at his quick answer and asked him how he did it.
the work is this.
1+2+3+---------+100
100+99+98+--------+1
-----------------------------------
101+101+101+.........+101 (100 times)
=10100
here the required sum is taken twice, so the sum is 10100/2=5050.
This method is only used to find the sum of n terms in any arithmetic series.
first term=a
second term=a+d where d is the common difference.
third tem=a+2d
.
.
n th term=a+(n-1)d
let the sum be S
taking the terms in reverse order and adding gives
2S=2a+(n-1)d *n
S=(n/2)*{2a+(n-1)d}
The sum of first n natural numbers can be obtained by substituting a=1 and d=1,
S=(n/2)*{2+(n-1)}=n(n+1)/2
when n=100, S=100*101/2=5050.

Answer 5

answer=100 X (101 )/2

=50 X 101

= 5050.

Answer 6

Series is in AP.

Formula for calculating this is n/2(first term + last term)

100/2(1+100)=5050

Answer 7

Sequentially adding the numbers 1 through 100 will return the product 5,050.

Another way to perform the action is to use the formula
x(x+1)/2=y, given x is the number of numbers being added -

100(100+1)/2 = 5,050

Answer 8

1+2+3+...+n={n*(n+1)}/2
so that is 100*101/2=5050
have some practice to prove the formula!!

Answer 9

[n(n+1)]/2
[100*101]/2
10100/2 = 5050

Answer 10

106.

But if you want me to fill in the blanks, 5050.

49+51 = 100
48+52 = 100
47+53 = 100
.
.
.
1+99 = 100

Then there's 50 itself and 100 itself.

Answer 11

the formula is : n(n+1)/2 where n is the total number of consecutive integers being added.
so the ans is 5050