## Introduction

In this C++ tutorial, I will show you how to find the sum of the series *1 +3 +5…*. This is a beginner level interview question. We will show you two different ways to solve it.

### Method 1 : Using a for loop :

Any problem that you need to iterate, you can use a *loop*. It can be a *for* loop, *while* loop or *do-while* loop.

For this problem, we need to find the sum of a series. The difference between two consecutive elements is *2*. We will use one loop, that will run from *1* to *total numbers*. We will use one separate variable to store the sum. For each iteration, we will increment this sum variable or we will add the current element to this sum variable.

```
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int total;
int sum = 1;
cout << "Enter total number : " << endl;
cin >> total;
for (int i = 2; i <= total; i++)
{
sum += 2 * i - 1;
}
cout << "Sum : " << sum << endl;
}
```

Here, the *sum* variable is initialized as *1*. The loop runs from *i = 2* to *i = total*. It adds *2*i - 1* on each iteration.

Suppose, we are adding the first five elements of the series :

```
i = 1, current series element is = 2*1 - 1 = 1
i = 2, current series element is = 2*2 - 1 = 3
i = 3, current series element is = 2*3 - 1 = 5
i = 4, current series element is = 2*4 - 1 = 7
```

So, *2*i - 1* always results in the current series element. Once the loop will end, *sum* will hold the sum of all digits or the first *total* digits.

### Sample output :

```
Enter total number :
5
Sum : 25
Enter total number :
20
Sum : 400
```

### Using the arithmetic series formula :

*1 + 3 + 5….* is an arithmetic series with common difference *2*. We can find out the sum of arithmetic series using the below formula :

`(n/2)*[2a+(n−1)d]`

Where *n* is the total number of elements starting from first, *a* is the first element and *d* is a common difference. For our example, we have *a = 1* and *d = 2*. We can take the value of *n* from the user as input and calculate the sum :

```
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int n;
int sum;
cout << "Enter total number : " << endl;
cin >> n;
sum = (n/2)*(2 + (n-1)*2);
cout << "Sum : " << sum << endl;
}
```

### Output :

```
Enter total number :
10
Sum : 100
Enter total number :
15
Sum : 210
```

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