(n)( n+1)/2 where n = 50. (50 x 51)/2 = 2,525
(n - 2)x180 where n = number of sides so ... 51 x 180
51
The sum of n and 5 is n+ 5
51 + 57 = 108
n=9. Because n+42=51. Subtract 42 from both sides and n=9
Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275
Sum of 1st n even numbers: count*average = n * (2 + 2*n)/2 = n * (n+1) Sum = 50 * (2+100)/2 = 50*51 = 2550
To find the sum of all numbers from 51 to 150, we can use the formula for the sum of an arithmetic series: (n/2)(first term + last term), where n is the number of terms. In this case, the first term is 51, the last term is 150, and the number of terms is 150 - 51 + 1 = 100. Plugging these values into the formula, we get (100/2)(51 + 150) = 50 * 201 = 10,050. Therefore, the sum of all numbers from 51 to 150 is 10,050.
(n)( n+1)/2 where n = 50. (50 x 51)/2 = 2,525
(n - 2)x180 where n = number of sides so ... 51 x 180
51
38+51 = 89
int n, N;N = some even numberfor (n=2; n
The sum of n and 5 is n+ 5
It is: 10 plus 51 = 61
-42