Find sum of Series with n-th term as n^2 – (n-1)^2

We are given an integer n and n-th term in a series as expressed below: 

Tn = n2 - (n-1)2

We need to find S_n mod (10⁹ + 7), where S_n is the sum of all of the terms of the given series and, 

Sn = T1 + T2 + T3 + T4 + ...... + Tn

Examples: 

Input : 229137999
Output : 218194447

Input : 344936985
Output : 788019571

Let us do some calculations, before writing the program. T_n can be reduced to give 2n-1 . Let’s see how: 

Given, Tn = n2 - (n-1)2
Or, Tn =  n2 - (1 + n2 - 2n)
Or, Tn =  n2 - 1 - n2 + 2n
Or, Tn =  2n - 1. 

Now, we need to find ?T_n. 

?T_n = ?(2n – 1)

We can simplify the above formula as, 
?(2n – 1) = 2*?n – ?1 
Or, ?(2n – 1) = 2*?n – n. 
Where, ?n is the sum of first n natural numbers. 
We know the sum of n natural number = n(n+1)/2.
Therefore, putting this value in the above equation we will get, 

?T_n = (2*(n)*(n+1)/2)-n = n²

Now the value of n² can be very large. So instead of direct squaring n and taking mod of the result. We will use the property of modular multiplication for calculating squares: 

(a*b)%k = ((a%k)*(b%k))%k

Below is the implementation of above idea: 

CPP

// CPP program to find sum of given
// series.
#include<bits/stdc++.h>
using namespace std;

#define mod 1000000007

// Function to find sum of series
// with nth term as n^2 - (n-1)^2
long long findSum(long long n)
{
    return ((n%mod)*(n%mod))%mod;
}

// Driver code
int main()
{
    long long n = 229137999;
    cout << findSum(n);
    return 0;
}

Java

// Java program to find sum of given
// series.

public class FINDSUM {

    static long mod = 1000000007;

    public static long findSum(long n)
    {
        return ((n % mod) * (n % mod)) % mod;
    }

    public static void main(String[] args) {
        long n = 229137999;
        System.out.print (findSum(n));
    }
}

// Contributed by _omg

Python 3

# Python program to find sum of given
# series.

mod = 1000000007
def findSum(n):
    return ((n % mod) * (n % mod)) % mod


# main()
n = 229137999
print (findSum(n))

# Contributed by _omg

C#

// C# program to find sum of given
// series.
using System;

class GFG
{
    static long mod = 1000000007;

    static long findSum(long n)
    {
        return ((n % mod) * (n % mod)) % mod;
    }

    public static void Main()
    {
        long n = 229137999;
        Console.Write (findSum(n));
    }
}

// This code is contributed by _omg

PHP

<?php
// PHP program to find
// sum of givenseries.

$mod = 1000000007;

// Function to find sum of series
// with nth term as n^2 - (n-1)^2
function findSum($n)
{
    global $mod;
    return (($n % $mod) *
            ($n % $mod)) % $mod;
}

// Driver code
$n = 229137999;
echo(findSum($n));

// This code is contributed by Ajit.
?>

Javascript

<script>

    // Javascript program to find sum of given series.

    let mod = 1000000007;

    function findSum(n)
    {
        return ((n % mod) * (n % mod)) % mod + 1;
    }

    let n = 229137999;
      document.write(findSum(n));

</script>

Output:

218194447

Time Complexity: O(1) 
Auxiliary Space: O(1)