Given a quadratic series as given below, the task is to find sum of first n terms of this series.

Sn = 3 + 7 + 13 + 21 + 31 + ….. + upto n terms

**Examples:**

Input: N = 3 Output: 23 Input: N = 4 Output: 44

**Approach:**

Let the series be represented as

S_{n}= 3 + 7 + 13 + ... + t_{n}

where

**S**represents the sum of the series till n terms._{n}**t**represents the nth term of the series._{n}

Now, to formulate the series, the elements need to be formed by taking the difference of the consecutive elements of the series.

Equation 1: Sn = 3 + 7 + 13 + 21 + 31 +…..+ tn-1 + tn

Equation 2: Sn = 0 + 3 + 7 + 13 + 21 + 31 + …… + tn-1 + tn

(writing the above series by shifting all elements to right by 1 position)

Now, Subtract Equation 2 from Equation 1 i.e. **(Equation 1 – Equation 2)**

Sn – Sn = (3 – 0) + (7 – 3) + (13 – 7) + (31 – 21) + …… + (tn- tn-1) – tn

=> 0 = 3 + 4 + 6 + 8 + 10 + …… + (tn – tn-1) – tn

In above series, leaving 3 aside, terms starting from 4 to (tn – tn-1) will form an A.P.

Since formula of sum of n terms of A.P. is:

Sn = n*(2*a + (n – 1)*d)/2

which implies,

In series: 4 + 6 + 8 + … + (tn – tn-1)

AP is formed with (n-1) terms.

Hence,

Sum of this series: (n-1)*(2*4 + (n-2)*2)/2

Therefore, the original series:

0 = 3 + (n-1)*(2*4 + (n-2)*2)/2 – tn

where **tn = n^2 + n + 1** which is the nth term.

Therefore,

Sum of first n terms of series will be:

tn = n^2 + n + 1

Sn = (n^2) + n + (1)

Sn = n*(n+1)*(n+2)/6 + n*(n+1)/2 + n

Sn = n*(n^2 + 3*n + 5)/3

**Below is the implementation of above approach:**

## C++

`// C++ program to find sum of first n terms` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `calculateSum(` `int` `n)` `{` ` ` `// Sum = n*(n^2 + 3*n + 5)/3` ` ` `return` `n * (` `pow` `(n, 2) + 3 * n + 5) / 3;` `}` `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` `import` `java.util.*;` `class` `solution` `{` `//function to calculate sum of n terms of the series` `static` `int` `calculateSum(` `int` `n)` `{` ` ` `// Sum = n*(n^2 + 3*n + 5)/3` ` ` `return` `n * (` `int` `) (Math.pow(n, ` `2` `) + ` `3` `* n + ` `5` `)/ ` `3` `;` `}` `public` `static` `void` `main(String arr[])` `{` ` ` `// number of terms to be included in the sum` ` ` `int` `n = ` `3` `;` ` ` `// find the Sum` ` ` `System.out.println(` `"Sum = "` `+calculateSum(n));` `}` `}` |

## Python3

`# Python 3 program to find sum` `# of first n terms` `from` `math ` `import` `pow` `def` `calculateSum(n):` ` ` ` ` `# Sum = n*(n^2 + 3*n + 5)/3` ` ` `return` `n ` `*` `(` `pow` `(n, ` `2` `) ` `+` `3` `*` `n ` `+` `5` `) ` `/` `3` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# number of terms to be included` ` ` `# in the sum` ` ` `n ` `=` `3` ` ` `# find the Sum` ` ` `print` `(` `"Sum ="` `, ` `int` `(calculateSum(n)))` `# This code is contributed by` `# Sanjit_Prasad` |

## C#

`// C# program to find sum of first n terms` `using` `System;` `class` `gfg` `{` ` ` `public` `double` `calculateSum(` `int` `n)` ` ` `{` ` ` `// Sum = n*(n^2 + 3*n + 5)/3` ` ` `return` `(n * (Math.Pow(n, 2) + 3 * n + 5) / 3);` ` ` `}` `}` `//driver code` `class` `geek` `{` ` ` `public` `static` `int` `Main()` ` ` `{` ` ` `gfg g = ` `new` `gfg();` ` ` `// number of terms to be included in the sum` ` ` `int` `n = 3;` ` ` `//find the Sum` ` ` `Console.WriteLine( ` `"Sum = {0}"` `, g.calculateSum(n));` ` ` `return` `0;` ` ` `}` `}` |

## PHP

`<?php` `// PHP program to find sum` `// of first n terms` `function` `calculateSum(` `$n` `)` `{` ` ` `// Sum = n*(n^2 + 3*n + 5)/3` ` ` `return` `$n` `* (pow(` `$n` `, 2) + 3 *` ` ` `$n` `+ 5) / 3;` `}` `// Driver Code` `// number of terms to be` `// included in the sum` `$n` `= 3;` `// find the Sum` `echo` `"Sum = "` `. calculateSum(` `$n` `);` `// This code is contributed by mits` `?>` |

## Javascript

`<script>` `// Javascript program to find sum of first n terms` `// Function to find the quadratic` `// equation whose roots are a and b` `function` `calculateSum(n)` `{` ` ` ` ` `// Sum = n*(n^2 + 3*n + 5)/3` ` ` `return` `n * (Math.pow(n, 2) + 3 * n + 5 ) / 3;` `}` `// Driver Code` `// Number of terms to be` `// included in the sum` `var` `n = 3;` `// Find the Sum` `document.write(` `"Sum = "` `+ calculateSum(n));` `// This code is contributed by Ankita saini` ` ` `</script>` |

**Output:**

Sum = 23

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