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Find sum of Series with n-th term as n^2 – (n-1)^2
  • Last Updated : 03 May, 2021

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 Sn mod (109 + 7), where Sn 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. Tn 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 ∑Tn
 

∑Tn = ∑(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, 
 

∑Tn = (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

 

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