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

• Last Updated : 12 Jul, 2022

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 = 3
Output: 4.5

Input: N = 4
Output: 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 N terms in AP.

The Sum of N terms of an AP can also be given by SN,

SN = N * [First term + Last term] / 2

SN = 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-

SN = N * (2 + (N – 1) * 0.5) / 2

Illustration:

Input: N = 5
Output: 10
Explanation:
SN = 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 ``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

 ``

Output

`13.5`

Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.

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