# Series summation if T(n) is given and n is very large

Given a sequence whose **n ^{th}** term is

T(n) = n

^{2}– (n – 1)^{2}

The task is to evaluate the sum of first n terms i.e.

S(n) = T(1) + T(2) + T(3) + … + T(n)

Print **S(n) mod (10 ^{9} + 7)**.

**Examples:**

Input:n = 3

Output:9

S(3) = T(1) + T(2) + T(3) = (1^{2}– 0^{2}) + (2^{2}– 1^{2}) + (3^{2}– 2^{2}) = 1 + 3 + 5 = 9

Input:n = 10

Output:100

**Approach:** If we try to find out some initial terms of the sequence by putting **n = 1, 2, 3, …** in **T(n) = n ^{2} – (n – 1)^{2}**, we find the sequence

**1, 3, 5, …**

Hence, we find an A.P. where

**first term**is

**1**and

**d**(common difference between consecutive

terms) is

**2**.

The formula for the sum of n terms of A.P is

S(n) = n / 2 [ 2 * a + (n – 1) * d ]

where **a** is the first term.

So, putting **a = 1** and **d = 2**, we get

S(n) = n

^{2}

.

Below is the implementation of above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` `#define ll long long int ` `#define MOD 1000000007 ` ` ` `// Function to return the sum ` `// of the given series ` `int` `sumOfSeries(` `int` `n) ` `{ ` ` ` `ll ans = (ll)` `pow` `(n % MOD, 2); ` ` ` ` ` `return` `(ans % MOD); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 10; ` ` ` `cout << sumOfSeries(n); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `class` `GFG ` `{ ` ` ` `public` `static` `final` `int` `MOD = ` `1000000007` `; ` ` ` `// Function to return the sum ` `// of the given series ` `static` `int` `sumOfSeries(` `int` `n) ` `{ ` ` ` `int` `ans = (` `int` `)Math.pow(n % MOD, ` `2` `); ` ` ` ` ` `return` `(ans % MOD); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `n = ` `10` `; ` ` ` `System.out.println(sumOfSeries(n)); ` `} ` `} ` ` ` `// This code is contributed by Code_Mech. ` |

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## Python3

`# Python 3 implementation of the approach ` `from` `math ` `import` `pow` ` ` `MOD ` `=` `1000000007` ` ` `# Function to return the sum ` `# of the given series ` `def` `sumOfSeries(n): ` ` ` `ans ` `=` `pow` `(n ` `%` `MOD, ` `2` `) ` ` ` ` ` `return` `(ans ` `%` `MOD) ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `n ` `=` `10` ` ` `print` `(` `int` `(sumOfSeries(n))) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `const` `int` `MOD = 1000000007; ` ` ` `// Function to return the sum ` `// of the given series ` `static` `int` `sumOfSeries(` `int` `n) ` `{ ` ` ` `int` `ans = (` `int` `)Math.Pow(n % MOD, 2); ` ` ` ` ` `return` `(ans % MOD); ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main() ` `{ ` ` ` `int` `n = 10; ` ` ` `Console.Write(sumOfSeries(n)); ` `} ` `} ` ` ` `// This code is contributed ` `// by Akanksha Rai ` |

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## PHP

`<?php ` `// PHP implementation of the approach ` `$GLOBALS` `[` `'MOD'` `] = 1000000007; ` ` ` `// Function to return the sum ` `// of the given series ` `function` `sumOfSeries(` `$n` `) ` `{ ` ` ` `$ans` `= pow(` `$n` `% ` `$GLOBALS` `[` `'MOD'` `], 2); ` ` ` ` ` `return` `(` `$ans` `% ` `$GLOBALS` `[` `'MOD'` `]); ` `} ` ` ` `// Driver code ` `$n` `= 10; ` `echo` `sumOfSeries(` `$n` `); ` ` ` `// This code is contributed by Ryuga ` `?> ` |

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**Output:**

100

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