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What is the sum of the first 50 natural numbers?
sum of n natural number is n(n+1)/2 first 50 number sum is 50(50+1)/2 = 1275
What is the sum of odd numbers from 1 to 100?
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
What are the 9 odd numbers whose sum is 50 from 1 to 50?
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.
What is the sum of 50 greatest negative integers?
i got the same question for homework in functions. tn=a+(n-1)d -50=-1+(n-1)-1 -50=-1-1n+1 -50=-1n therefore n=50 plug that into your sum equation...for convenience i used Sn=50/2 (-1-50) Sn=25 (-51) Sn= -1275 hope this helps!
The sum of two numbers is 65 their product of these two is 750 what is the sum of the reciprocals of the two numbers?
(50 + 15) = 65 (50 x 15) = 750 1/50 + 1/15 = 13/150
