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

Given a sequence whose nth term is

T(n) = n2 – (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 (109 + 7).

Examples:

Input: n = 3
Output: 9
S(3) = T(1) + T(2) + T(3) = (12 – 02) + (22 – 12) + (32 – 22) = 1 + 3 + 5 = 9

Input: n = 10
Output: 100

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: If we try to find out some initial terms of the sequence by putting n = 1, 2, 3, … in T(n) = n2 – (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) = n2

.

Below is the implementation of above approach:

 `// C++ implementation of the approach ` `#include ` `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; ` `} `

 `// 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. `

 `# 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 `

 `// 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 `

 ` `

Output:
```100
```

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