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

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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:
S(3) = T(1) + T(2) + T(3) = (12 – 02) + (22 – 12) + (32 – 22) = 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) = 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++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll 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




// 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.


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


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


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
?>


Javascript




<script>
 
// javascript program for the above approach
 
let MOD = 1000000007;
 
// Function to return the sum
// of the given series
function sumOfSeries(n)
{
    let ans = Math.pow(n % MOD, 2);
 
    return (ans % MOD);
}
 
// Driver Code
     
        let n = 10;
    document.write(sumOfSeries(n));
 
</script>


Output

100

Time Complexity: O(1) because calculation square of a number using pow function takes constant time.
Auxiliary Space: O(1), since no extra space has been taken.



Last Updated : 03 Nov, 2022
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