The sum of the first 50 whole numbers is 1,225.
the sum of first 50 consecutive odd numbers is 9801
The sum of the first 50 counting numbers, excluding zero, is 1,251.
sum of n natural number is n(n+1)/2 first 50 number sum is 50(50+1)/2 = 1275
To find the 9 odd numbers whose sum is 50 from 1 to 50, we can first calculate the sum of all odd numbers from 1 to 50, which is (50^2)/2 = 625. Next, we subtract the sum of the first 9 odd numbers (1+3+5+7+9+11+13+15+17 = 81) from the total sum, resulting in 625 - 81 = 544. Therefore, the 9 odd numbers whose sum is 50 from 1 to 50 are 19, 21, 23, 25, 27, 29, 31, 33, and 36.
Sum of 1st n even numbers: count*average = n * (2 + 2*n)/2 = n * (n+1) Sum = 50 * (2+100)/2 = 50*51 = 2550
The sum of the first 50 odd numbers is 2,500.
The sum of the first 50 even numbers is 2,550.
The sum of the first 50 natural numbers is 1,251.
the sum of first 50 consecutive odd numbers is 9801
The sum of the first 50 counting numbers, excluding zero, is 1,251.
sum of n natural number is n(n+1)/2 first 50 number sum is 50(50+1)/2 = 1275
of all the things i got is 2550
2500
To find the 9 odd numbers whose sum is 50 from 1 to 50, we can first calculate the sum of all odd numbers from 1 to 50, which is (50^2)/2 = 625. Next, we subtract the sum of the first 9 odd numbers (1+3+5+7+9+11+13+15+17 = 81) from the total sum, resulting in 625 - 81 = 544. Therefore, the 9 odd numbers whose sum is 50 from 1 to 50 are 19, 21, 23, 25, 27, 29, 31, 33, and 36.
Sum of 1st n even numbers: count*average = n * (2 + 2*n)/2 = n * (n+1) Sum = 50 * (2+100)/2 = 50*51 = 2550
Why, that's just 6 times the sum of the numbers from 1 to 50 !The sum of the numbers from 1 to 50 is:1 + 50 = 512 + 49 = 513 + 48 = 514 + 47 = 515 + 46 = 51...etc. There are 25 pairs, and each pair sums to 51. So the sum of 1 ... 50 is (25 x 51) = 1,275.The sum of the first 50 multiples of 6 is (1,275 x 6) = 7,650.
51