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Assuming you mean the first n counting numbers then:

let S{n} be the sum; then:

S{n} = 1 + 2 + ... + (n-1) + n

As addition is commutative, the sum can be reversed to give:

S{n} = n + (n-1) + ... + 2 + 1

Now add the two versions together (term by term), giving:

S{n} + S{n} = (1 + n) + (2 + (n-1)) + ... + ((n-1) + 2) + (n + 1)

→ 2S{n} = (n+1) + (n+1) + ... + (n+1) + (n+1)

As there were originally n terms, this is (n+1) added n times, giving:

2S{n} = n(n+1)

→ S{n} = ½n(n+1)

The sum of the first n counting numbers is ½n(n+1).

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The sum of the first n counting numbers is n*(n+1)/2

Q: What is the sum of the first n numbers?

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n*(n+1)=sum 100*(100+1)=10100

The sum of the first N square numbers is: N*(N+1)*(2N+1)/6 So putting N = 20 gives 2870.

Sum of first n natural numbers is (n) x (n + 1)/2 Here we have the sum = 100 x (101)/2 = 50 x 101 = 5050

To find the sum of the numbers, we must first know the value of n. This was not included in your question.

Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275

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The sum of the first n cubed numbers is: [n*(n+1)/2]2 which is the same as the square of the sum of the first n numbers.

int sum = 0; int i; for(i = 0; i < n; ++i) { sum += i; }

The sum of the first "n" numbers is equal to n(n+1)/2.

1 Sum of first n natural numbers = n(n+1)2[Formula.]2 Arthmetic mean of first n natural numbers = Sum of the numbers n[Formula.]3 = n(n+1)2n = n+124 So, the Arthmetic mean of first n natural numbers = n+12

The sum of the first n natural numbers is n*(n+1)/2 There are n numbers so their mean = (n+1)/2

n*(n+1)

To get a list of the squares of the first 1000 numbers we can do:> [n^2 | n sum [n^2 | n

n*(n+1)=sum 100*(100+1)=10100

The sum of the first N square numbers is: N*(N+1)*(2N+1)/6 So putting N = 20 gives 2870.

Sum = n/2[2Xa1+(n-1)d] where n is last number, a1 is the first number & d is the common difference between the numbers, here d=2 for the even /odd numbers. Sum = n/2 [2Xa1+(n-1)2]

Sum of first n natural numbers is (n) x (n + 1)/2 Here we have the sum = 100 x (101)/2 = 50 x 101 = 5050

To find the sum of the numbers, we must first know the value of n. This was not included in your question.