# Find the sum of the series 1+(1+2)/2+(1+2+3)/3+… till N terms

Given a number **N**, the task is to find the sum of the below series till **N** terms.

1+(1+2)/2+(1+2+3)/3+… till N terms

**Examples:**

Input:N = 3Output:4.5

Input:N = 4Output:7

**Approach: **

From the given series, find the formula for the **Nth** term:

1st term = 1 = 1

2nd term = (1+2)/2 = 1.5

3rd term = (1+2+3)/3 = 2

4th term = (1+2+3+4)/4 = 2.5

.

.

Nth term = N*(N +1)/(2*N) = (N+1)/2

**Derivation:**

The following series of steps can be used to derive the formula to find the sum of N terms-

The sequence-

1+(1+2)/2+(1+2+3)/3+… till N terms

can be written as-

1 +1.5 +2 +2.5 +3 +3.5 +4 +…till N terms

The above series is an Arithmetic Progression(AP) series. So, we can directly apply the formula of sum of

Nterms in AP.The Sum of

Nterms of an AP can also be given byS,_{N}S

_{N}= N * [First term + Last term] / 2S

_{N}= N * [2 * a + (N – 1) * d] / 2 -(1)From the above equation, it is known that,

a(first term)=1, d(common difference) = 1.5 -1= 0.5

Substituting the values of a and d in the equation (1), we get-

S

_{N}= N * (2 + (N – 1) * 0.5) / 2

**Illustration:**

Input:N = 5Output:10Explanation:

S_{N}= 5 * [2 * 1 + (5 – 1) * 0.5] / 2

= 5 * (2 + 2) / 2

= 5 * 2

= 10

Below is the C++ program to implement the above approach:

## C++

`// C++ program to find the sum of the` `// series 1+(1+2)/2+(1+2+3)/3+...` `// till N terms` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the sum` `// upto N term of the series` `double` `sumOfSeries(` `int` `N)` `{` ` ` `return` `((` `double` `)N` ` ` `* (2 + ((` `double` `)N - 1) * 0.5))` ` ` `/ 2;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Get the value of N` ` ` `int` `N = 6;` ` ` `cout << sumOfSeries(N);` ` ` `return` `0;` `}` |

## Java

`// Java program to find the sum of the` `// series 1+(1+2)/2+(1+2+3)/3+...` `// till N terms` `import` `java.util.*;` `public` `class` `GFG` `{` ` ` `// Function to return the sum` ` ` `// upto N term of the series` ` ` `static` `double` `sumOfSeries(` `int` `N)` ` ` `{` ` ` `return` `((` `double` `)N` ` ` `* (` `2` `+ ((` `double` `)N - ` `1` `) * ` `0.5` `))` ` ` `/ ` `2` `;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `// Get the value of N` ` ` `int` `N = ` `6` `;` ` ` `System.out.println(sumOfSeries(N));` ` ` `}` `}` `// This code is contributed by Samim Hossain Mondal.` |

## Python

`# Python program to find the sum of the` `# series 1+(1+2)/2+(1+2+3)/3+...` `# till N terms` `# Function to return the sum` `# upto N term of the series` `def` `sumOfSeries(N):` ` ` ` ` `return` `(N ` `*` `(` `2` `+` `(N ` `-` `1` `) ` `*` `0.5` `) ` `/` `2` `)` `# Driver Code` `# Get the value of N` `N ` `=` `6` `print` `(sumOfSeries(N))` `# This code is contributed by Samim Hossain Mondal.` |

## C#

`// C# program to find the sum of the` `// series 1+(1+2)/2+(1+2+3)/3+...` `// till N terms` `using` `System;` `class` `GFG` `{` ` ` `// Function to return the sum` ` ` `// upto N term of the series` ` ` `static` `double` `sumOfSeries(` `int` `N)` ` ` `{` ` ` `return` `((` `double` `)N` ` ` `* (2 + ((` `double` `)N - 1) * 0.5))` ` ` `/ 2;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `// Get the value of N` ` ` `int` `N = 6;` ` ` `Console.Write(sumOfSeries(N));` ` ` `}` `}` `// This code is contributed by Samim Hossain Mondal.` |

## Javascript

`<script>` ` ` `// JavaScript code for the above approach` ` ` `// Function to return the sum` ` ` `// upto N term of the series` ` ` `function` `sumOfSeries(N) {` ` ` `return` `(N` ` ` `* (2 + (N - 1) * 0.5))` ` ` `/ 2;` ` ` `}` ` ` `// Driver Code` ` ` `// Get the value of N` ` ` `let N = 6;` ` ` `document.write(sumOfSeries(N));` ` ` `// This code is contributed by Potta Lokesh` ` ` `</script>` |

**Output**

13.5

**Time Complexity: **O(1) **Auxiliary Space:** O(1)