what pattern i should use to add 1+2+3+4+5+6+...+98+99+100...please help me,thanks?

Answer 1

Pair up 1 and 100, 2 and 99, etc. Then you have 50 pairs that all add to 101.

Answer 2

98+2+100+99+1+6+4+3+5

Answer 3

1+2+3+4+5+6+...+98+99+100 =
(1+100)+(2+99)+(3+98)+........ + ( 50 +51) =
101+101+101+....+101 =
101·50 = 5050


1+2+3++.....+n = (1+n) · n/2

Saludos.

Answer 4

(1+100)*100/2=5050.

There are 50 of 101... 1+100, 2+99, 3+98, etc.

Answer 5

You should use the formula for the Sum to n terms of an arithmetic progression.
This is :
Sn= 0.5n [2a + (n-1)d]
where n stands for the nth term of the progression,
and a stands for the first term of the progression
and d stands for the difference between any two consecutive terms in the progression

In this case, you have 100 terms in your progression, so n =100
And a = 1
And d = 1
Therefore,
S100 = (0.5 x 100)[2(1) + (100 - 1)1]

= 50 x 101

= 5050

Answer 6

this is sum of n natural numbers
let S = sum
S = 1 + 2 + 3 + ------- (n -2) + (n - 1) + n --- eqn (1)
S = n+ (n - 1)+(n - 2) + ----- 3 + 2 + 1 -- eqn (2)(by
writing the sum in reverse order

add eqn (1) and (2)
2S = (n+1)+(n+1)+(n + 1)-----------
in right hand side each term is (n+1) and there are n terms
so 2S = n(n+1)
S = n(n + 1)/2
if you want to find out the sum of first 100 numbers
substitute n = 100
S = 100(101)/2 = 50*101 = 5050

Answer 7

101 x 50 equals 5050, becuase 1 +100 equals 101, and 2 + 99 equals 101, etc. and there are 50 pairs of numbers, thus the above answer.

Answer 8

just multiply the last number by the next number then divide by 2, that is (100 x 101) / 2 = 5 050

Answer 9

(0)+1+2+3+4+5+6+...+98+99+100
100+0=100
99+1=100
98+2=100
97+3=100
and so on... see how it works?

Answer 10

you have several sets of 100.
100+0=100
99+1=100
98+2=100
etc. all the way to 51+49=100
so you have 50 sets of 100 and 50 left over
50(100) +50=
5000 +50= 5,050