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int sum = 0;

int i;

for(i = 0; i < n; ++i) {

sum += i;

}

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15y ago
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Wiki User

13y ago

n+(n-1)+(n-2)...

* * * * *

=n*(n + 1)/2

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14y ago

The first prime numbers are 2 and 3. Their sum is 5.

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MARYAM MULLA

Lvl 2
2y ago

The sum of the first prime numbers is 5

Because the first prime numbers are 2 & 3.

Thus , 2 + 3 =5

The sum is 5

I hope you understand my reason for my answer.

By grade 4 th student

byeee

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Wiki User

13y ago

it is n(n+1)/2

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Q: What is the Sum of first n natural numbers?
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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


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main() { int i, n, sum=0; cout&lt;&lt;"Enter the limit"; cin&gt;&gt;n; for(i=0;i&lt;=n;i++) sum=sum+i; cout&lt;&lt;"sum of "&lt;&lt;n&lt;&lt;" natural numbers ="&lt;&lt;sum; getch(); }


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


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Program to calculate the sum of first ten natural numbers?

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How do you calculate the sum of all numbers from 1 through 61?

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What is the sum of two consecutive natural numbers is 525?

Let n be an arbitrary natural number. The next larger natural number is n + 1. The sum of these natural numbers is n + (n + 1) n + (n + 1) = 525 2n + 1 = 525 2n = 524 n = 262 n + 1 = 263. The two natural numbers are 262 and 263.