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# Sum of the first N terms of the series 2,10, 30, 68,….

• Last Updated : 31 May, 2022

Given a number N, the task is to find the sum of first N terms of the below series:

Sn = 2 + 10 + 30 + 68 + … upto n terms

Examples:

```Input: N = 2
Output: 12
2 + 10
= 12

Input: N = 4
Output: 40
2 + 10 + 30 + 68
= 110```

Approach: Let, the nth term be denoted by tn.
This problem can easily be solved by splitting each term as follows :

```Sn = 2 + 10 + 30 + 68 + ......
Sn = (1+1^3) + (2+2^3) + (3+3^3) + (4+4^3) +......
Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^3 + 2^3 + 3^3 + 4^3 + ...upto n terms)```

We observed that Sn can broken down into summation of two series.
Hence, the sum of first n terms is given as follows:

```Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^3 + 2^3 + 3^3 + 4^3 + ...upto n terms)
Sn = n*(n + 1)/2 + (n*(n + 1)/2)^2```

Below is the implementation of above approach:

## C++

 `// C++ program to find sum of first n terms` `#include ` `using` `namespace` `std;`   `// Function to calculate the sum` `int` `calculateSum(``int` `n)` `{`   `    ``return` `n * (n + 1) / 2 ` `           ``+ ``pow``((n * (n + 1) / 2), 2);` `}`   `// Driver code` `int` `main()` `{` `    ``// number of terms to be` `    ``// included in the sum` `    ``int` `n = 3;`   `    ``// find the Sum` `    ``cout << ``"Sum = "` `<< calculateSum(n);`   `    ``return` `0;` `}`

## Java

 `// Java program to find sum of first n terms `   `public` `class` `GFG {` `    `  `    ``// Function to calculate the sum ` `    ``static` `int` `calculateSum(``int` `n) ` `    ``{ ` `      `  `        ``return` `n * (n + ``1``) / ``2`  `               ``+ (``int``)Math.pow((n * (n + ``1``) / ``2``), ``2``); ` `    ``} ` `    `  `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``// number of terms to be ` `        ``// included in the sum ` `        ``int` `n = ``3``; ` `      `  `        ``// find the Sum ` `        ``System.out.println(``"Sum = "``+ calculateSum(n)); ` `    ``}` `    ``// This Code is contributed by ANKITRAI1` `}`

## Python3

 `# Python program to find sum ` `# of first n terms`   `# Function to calculate the sum` `def` `calculateSum(n):` `    ``return` `(n ``*` `(n ``+` `1``) ``/``/` `2` `+` `        ``pow``((n ``*` `(n ``+` `1``) ``/``/` `2``), ``2``))`   `# Driver code`   `# number of terms to be` `# included in the sum` `n ``=` `3`   `# find the Sum` `print``(``"Sum = "``, calculateSum(n))`   `# This code is contributed by` `# Sanjit_Prasad`

## C#

 `// C# program to find sum of first n terms` `using` `System;` `class` `gfg` `{` `    ``// Function to calculate the sum` `    ``public` `void` `calculateSum(``int` `n)` `    ``{` `        ``double` `r = (n * (n + 1) / 2 + ` `                ``Math.Pow((n * (n + 1) / 2), 2));` `        ``Console.WriteLine(``"Sum = "` `+ r);` `    ``}`   `    ``// Driver code` `    ``public` `static` `int` `Main()` `    ``{` `        ``gfg g = ``new` `gfg();`   `        ``// number of terms to be` `        ``// included in the sum` `        ``int` `n = 3;` `        `  `        ``// find the Sum` `        ``g.calculateSum(n);` `        ``Console.Read();` `        ``return` `0;` `    ``}` `}`

## PHP

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## Javascript

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Output:

`Sum = 42`

Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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