51
(n)( n+1)/2 where n = 50. (50 x 51)/2 = 2,525
rearrange 1 to 50 so that you see this pattern: 50 = 50 1+49 = 50 2+48 = 50 ... 24+26 = 50 25 = 25 then add this up: 25 x 50 + 25 = 1275
1/2, or 50%.1/2, or 50%.1/2, or 50%.1/2, or 50%.
It is instructive to note that you can rewrite 1+2+3+4....+99+100 as 1+100+2+99+3+98....+50+51 Then you can group each pair of numbers as (1+100)+(2+99)+(3+98)+...+(50+51) All of these couplets add up to 101, and there are 50 of them. Thus the sum of the first 100 whole numbers is 50x101. This can be calculated at 5,050.
sum of n natural number is n(n+1)/2 first 50 number sum is 50(50+1)/2 = 1275
51
Using the formula for the sum of an arithmetic sequence;Sum = 1/2*n [n is the total number of terms] *(2a [a is the value of the first term] + (n-1) * d)[d is the common difference between each term]So the formula looks like this: Sum = 1/2*n (2a+(n-1)*d)n = 50 because there's 50 odd numbers in a hundred and 50 even numbers (50+50=100)a = 1d = 2So inputting the values gives:Sum= 1/2*50 (2*1+(50-1)* 2)Sum= 25*(2+49*2)Sum= 25*(2+98)Sum= 25*(100)Sum= 2500Answer= 2,500
Here are the factor pairs of 50:1 and 502 and 255 and 10Now add them up: 1 + 50 + 2 + 25 + 5 + 10 = 93
Sum of 1st n even numbers: count*average = n * (2 + 2*n)/2 = n * (n+1) Sum = 50 * (2+100)/2 = 50*51 = 2550
(n)( n+1)/2 where n = 50. (50 x 51)/2 = 2,525
1 + 2 + 5 + 10 + 25 + 50 = 93
Sum of numbers from 1 to 100:(1+100) + (2+99) + (3+98) + (4+97) + ... + (49+52) + (50+51) =50 pairs, each pair adds up to 101= (50 x 101)= 5,050
rearrange 1 to 50 so that you see this pattern: 50 = 50 1+49 = 50 2+48 = 50 ... 24+26 = 50 25 = 25 then add this up: 25 x 50 + 25 = 1275
1/2, or 50%.1/2, or 50%.1/2, or 50%.1/2, or 50%.
Sum of first n natural numbers is (n) x (n + 1)/2 Here we have the sum = 100 x (101)/2 = 50 x 101 = 5050
Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275