1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = (20 x (20 + 1))/2 = 210
The sum of the first N square numbers is: N*(N+1)*(2N+1)/6 So putting N = 20 gives 2870.
The sum of -5 and 20 is 15. The distance between -5 and 20 is 25. The sum of the numbers between -5 and 20 is 180. The sum of the numbers from -5 to 20 is 195.
Avreage = Sum/number of values So 10 = sum/2 Multiply both sides by 2: 20 = sum
The sum is the result you get when you add two or more numbers together. A single number can be the sum of two or more other numbers when they are added together but cannot be expressed as you ask in your question. Though you could ask "What numbers have the sum of 20?" For instance 20 is the sum of (10 + 10), (5 + 15), (2 + 7 + 11) etc., but is not in itself a sum.
1/x + 1/y y/xy + x/xy (y+x)/xy= 10/20 10/20=1/2 1/2 is the answer.
20 and 1. 20 + 1 = 21 20 - 1 = 19
Well, honey, the sum of counting numbers from 1 to 20 is 210. You add up 1 + 2 + 3... all the way to 20 and you get your answer. So, go ahead and impress your friends with your newfound math knowledge.
To write a program in QBasic that prints the sum of the even numbers from 1 to 20 in reverse order, you can follow these steps: DIM sum AS INTEGER sum = 0 FOR i = 20 TO 2 STEP -2 sum = sum + i NEXT i PRINT "The sum of even numbers from 1 to 20 is: "; sum This program initializes the sum to zero, iterates from 20 down to 2 in steps of -2 (to capture even numbers), adds each even number to the sum, and finally prints the result.
(1 + 20) x (20/2) ie 210
The sum of the squares of the first 20 natural numbers 1 to 20 is 2,870.
The sum of the first N square numbers is: N*(N+1)*(2N+1)/6 So putting N = 20 gives 2870.
It is 75.
1 and 20.
The sum of -5 and 20 is 15. The distance between -5 and 20 is 25. The sum of the numbers between -5 and 20 is 180. The sum of the numbers from -5 to 20 is 195.
Suppose the numbers are x and y Then the sum of their reciprocals is 1/x + 1/y = y/xy + x/xy = (y+x)/xy = 10/20 = 1/2
It is 1.
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