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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).
What else can I help you with?
What is the sum of the first 100 even numbers?
n*(n+1)=sum 100*(100+1)=10100
What is the sum of the squares of the first 20 natural numbers 1 to 20?
The sum of the first N square numbers is: N*(N+1)*(2N+1)/6 So putting N = 20 gives 2870.
What is the sum of the numbers from 1 to 100?
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
What is the sum of 27 and a number n?
To find the sum of the numbers, we must first know the value of n. This was not included in your question.
What is the sum of the numbers from 1 to 50?
Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275
How do you get the sum of consecutive cubed numbers?
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.
What is the Sum of first n natural numbers?
int sum = 0; int i; for(i = 0; i < n; ++i) { sum += i; }
What is the Sum of the 1st 50 odd numbers?
The sum of the first 50 odd numbers is 2,500. This can be calculated using the formula for the sum of the first ( n ) odd numbers, which is ( n^2 ). For ( n = 50 ), the sum equals ( 50^2 = 2,500 ).
How do you calculate the sum of all numbers from 1 through 61?
The sum of the first "n" numbers is equal to n(n+1)/2.
The variance of first n natural numbers is?
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
How do you find the mean of first n natural numbers?
The sum of the first n natural numbers is n*(n+1)/2 There are n numbers so their mean = (n+1)/2
What is the formula to find out the sum of first n even numbers?
n*(n+1)
Find the programming code for calculating the sum of the squares of the first 1000 numbers in HASKELL?
To get a list of the squares of the first 1000 numbers we can do:> [n^2 | n sum [n^2 | n
What is the sum of the first 100 even numbers?
n*(n+1)=sum 100*(100+1)=10100
What is the sum of the squares of the first 20 natural numbers 1 to 20?
The sum of the first N square numbers is: N*(N+1)*(2N+1)/6 So putting N = 20 gives 2870.
What is the formula to find sum of n even numbers?
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]
What is the sum of the numbers from 1 to 100?
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
