c++ : inpunt an integer N. get the average of all odd number from 1 to N.

Question

i can't figure it out. thanks!

Answer 1

just use a for loop that starts from 1 and increses in +2 steps. add them all together and divide it by the amount of loops

Answer 2

if n is odd ... then output (n+1)/2
else output n/2
//☺👍

Answer 3

now see ☺
//now its correct in java as well as mathematically correct

Answer 4

useless comment because @gaurav corrected his comment

original comment:

let n = 5
sum of all odds is 1+3+5=9
average is 9/3= 3
output = 3

your formula: (5/2)+1= 3.5

3 != 3.5

correct me if im wrong

Answer 5

#include <iostream>
using namespace std;

int main() {
int x,y;
cout<<"input a number: ";
cin>>x;
for(int i=0;i<x;i++){
i=i+1;
y=i;
cout<<y<<endl;
}
return 0;
}

i know how to get the average literally.. but i dont know how to add all of the output number N then get the average.. thanks for the replies ;)

Answer 6

by the way I forgot to add, the problem is you don't need an if and else if statement.. that's y i cant fighre it out :(

Answer 7

you never mentioned java and i interpret formulas mathematically :p

Answer 8

wait, what exactly is it you cant do now?

Answer 9

i dont know how to add all the output number and get the average..can you help me complete the code? :)

Answer 10

double sum = 0;
int counter = 0;

for(int i = 1; i ≤ n; i+=2){
sum += i;
counter++;
}

double average = sum/counter;

or you use the formulars from above which gaurav provided