Algorithm to find the sum of first n natural numbers:1. Read n.2. Initialize N=1.3. Initialize sum S=0.4. Calculate S=S+N.5. Calculate N=N+1.6. If N>n, then goto step 7 else goto step 4.7. Write the sum S.8. Stop.
Well, you can look it up yourself in a table of prime numbers. But the general tendency is that, the higher you go, the less prime numbers you'll find in each interval. The long-term tendency is that among the first "n" numbers, you'll find n / ln(n) prime numbers, where ln(n) is the natural logarithm. This formula is not terribly accurate for small numbers, but it gets better and better as "n" gets larger.
The set of natural numbers or counting numbers N is a subset of the set of real numbers R. N = {1, 2, 3, ...) R = {..., -2, -1, -0.5, 0, 1, √2, 2, 3, π, ...}
The general equation to find the sum of the numbers 1 to n is: (n*(n+1))/2So, for n=10, you have:(10*(10+1))/2(10*11)/2110/255
Let n = smallest of the odd numbers, then let n+2 = the larger of the two numbers (Remember, 1 is not a prime number.) n+ n+2 = {(2)(7)}2 2n +2 = 142 2n = 196 -2 2n = 194 n = 97 n + 2 = 99
The sum of the first n natural numbers is n*(n+1)/2 There are n numbers so their mean = (n+1)/2
printf ("%g\n", (n+1.0)/2);
N stands for natural/normal. it was the first set of numbers ever classified, and was actually only classified after imaginary and complex numbers, coz before that, there was no need.
Formula for sum of first natural number = n(n+1)/2 , here n=225 so, answer is 225(225+1)/2 = 25425
The sum of the first N square numbers is: N*(N+1)*(2N+1)/6 So putting N = 20 gives 2870.
Mean = (sum of the n numbers)/n
#include<stdio.h> main(void) { int n; int i; printf("Type n: "); scanf("%d",&n); for(i=1;i<=n;i++) //generates natural numbers from 1,....,n. printf("%d\n",i); //prints these numbers to standard output }
I have found the coefficient of variation of the first natural numbers and also other functions.
int sum = 0; int i; for(i = 0; i < n; ++i) { sum += i; }
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
Algorithm to find the sum of first n natural numbers:1. Read n.2. Initialize N=1.3. Initialize sum S=0.4. Calculate S=S+N.5. Calculate N=N+1.6. If N>n, then goto step 7 else goto step 4.7. Write the sum S.8. Stop.
The average of the first n even numbers is n+1, so 101.The average of the first n even numbers is n+1, so 101.The average of the first n even numbers is n+1, so 101.The average of the first n even numbers is n+1, so 101.