Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + + (1+3+5+7+....+(2n-1)) in C++

In this problem, we are given an integer n. Our task is to create a program to find the sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + + (1+3+5+7+....+(2n-1)).

From this series, we can observe that ith term of the series is the sum of first i odd numbers.

Let’s take an example to understand the problem,

Input

n = 3

Output 

14

Explanation −(1) + (1+3) + (1+3+5) = 14

A simple solution to this problem is using a nested loop and then add all odd numbers to a sum variable. Then return the sum.

Example

Program to illustrate the working of our solution,

 Live Demo

#include <iostream>
using namespace std;
int calcSeriesSum(int n) {
   int sum = 0, element = 1;
   for (int i = 1; i <= n; i++) {
      element = 1;
      for (int j = 1; j <= i; j++) {
         sum += element;
         element += 2;
      }
   }
   return sum;
}
int main() {
   int n = 12;
   cout<<"Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + ... + (1+3+5+7+ ... + (2"<<n<<"-1)) is "<<calcSeriesSum(n);
   return 0;
}

Output

Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + ... + (1+3+5+7+ ... + (2*12-1)) is 650

This approach is not effective as it uses two nested loops.

A more efficient approach is to mathematically find the general formula to find the sum of the series.

Sum of n odd numbers,

= (1) + (1+3) + (1+3+5) + …. (1+3+5+... + 2n-1)

= n2

First, let’s see the sum of first n odd number, which represents the individual elements of the series.

Sum of series,

sum = (1) + (1+3) + (1+3+5) + … + (1+3+5+ … + 2n-1)
sum = ∑ (1+3+5+ … + 2n-1)
sum = ∑ n2
sum = [n * (n+1) * (2*n -1)]/6

Example

Program to illustrate the working of our solution,

 Live Demo

#include <iostream>
using namespace std;
int calcSeriesSum(int n) {
   return ( n*(n + 1)*(2*n + 1) )/6;
}
int main() {
   int n = 9;
   cout<<"Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + ... + (1+3+5+7+ ... + (2*"<<n<<"-1)) is "<<calcSeriesSum(n);
return 0;
}

Output

Sum of the series 1 + (1+3) + (1+3+5) + (1+3+5+7) + ... + (1+3+5+7+ ... + (2*9-1)) is 285