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There is a formula, called the formula for an arithmetic series, which you can use to calculate this: n/2 * [2a + (n - 1)d] Where: n is the number of terms (100) a is the first term (1) d is the difference between each term (1) Substituting these values we get: 100/2 * [2 + (100 - 1) * 1] = 50 * (2 + 99) = 5,050

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Q: What is the sum of 1 2 3 4 5 to 100?

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1+2+3+4.....+100 = 101*50= 5050

The sum of the numbers up to 100 is equal to (100 x 101)/2 = 5050

To solve for the sum of the first n number of cubes, use (1+2+3+4+5...+n)2. So for n=100, (1+2+3+4+5...+100)2 = c (cubes) (101 * 50)2 = c 50502 = c c = 25502500

1/2+1/2=1 1/3+1/3=2/3 1-2/3=1/3 Answer: 1/3

The sum of the positive integers between 99 and 999 is 299500 Since a = 100, and d = 2 note that the last term will be 998 using nth tem of an Ap Un= a+(n-1)d 998= 100+ (n-1)2 Then 998- 100= (n- 1)2 = 898= (n-1)2 449= n-1 n= 450 Sum of the terms = n/2(2a+(n-1)d) 450/2(2(100) + 449*2) Ans= 247050 =

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1+2+3+4.....+100 = 101*50= 5050

The sum of these composite numbers, including 100, is 3989.The sum of all numbers between 1 and 100 is 5,050. From this we need to subtract the sum of 1 plus all the prime numbers below 100. The sum of the primes is 1,060. Subtracting (1 + 1060) or 1,061 from 5,050 yields 3,989.

1 = 1 = 1^21+3 = 4 = 2^21+3+5 = 9 = 3^21+3+5+7 = 16 = 4^2Thus the pattern is that the sum if the first n odd numbers is n^2 or1+3+...+(2n-1) = n^2So, putting n = 100, gives the sum as 100^2 = 10000.

5050

The sum of the numbers up to 100 is equal to (100 x 101)/2 = 5050

To solve for the sum of the first n number of cubes, use (1+2+3+4+5...+n)2. So for n=100, (1+2+3+4+5...+100)2 = c (cubes) (101 * 50)2 = c 50502 = c c = 25502500

its 5035the summarian notation tells you that(sum of all #from 0 to a number'N'(sum of all #from 0 to N) = (n)+(n-1)+(n-2)+(n-3)+...+(2) +(1) or(sum of all #from 0 to N) = (1)+ (2) + (3) + (4)+...+(n-1)+(n)the two different sums are aligned by columns. now add the two colunms accordingly and you'll get2x(sum of all #from 0 to N)=(n+1)+(n+1)+...+(n+1) (n+1)is added n timesso2x(sum of all #from 0 to N) =n(n+1)(sum of all #from 0 to N) =n(n+1)/2so (sum of 5 to 100) = (sum of 0 to 100)- (sum of 0 to 5)=100(101)/2 - 5(6)/2 = 5050 - 15 = 5035

You add the numbers in a loop. Here is an example in Java:int sum = 0;for (int i = 1; i

The sum of prime numbers between 1 and 25 is 100. 2 + 3 + 5 + 7 + 11 +13 +17 + 19 + 23 = 100

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.

3 + 2 = 5. (sum) 3 - 2 = 1. (difference)

100/3-1, 100/3 and 100/3+1 that is, 99, 100 and 101.