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Program to print the sum of the given nth term

Given the value of the n. You have to find the sum of the series where the nth term of the sequence is given by:

Tn = n2 – ( n – 1 )2

Examples :

Input : 3
Output : 9

Explanation: So here the tern of the sequence upto n = 3 are: 
             1, 3, 5 And hence the required sum is = 1 + 3 + 5 = 9

Input : 6
Output : 36

Simple Approach
Just use a loop and calculate the sum of each term and print the sum.

C++

// CPP program to find summation of series
#include <bits/stdc++.h>
using namespace std;
  
int summingSeries(long n)
{
    // use of loop to calculate
    // sum of each term
    int S = 0; 
    for (int i = 1; i <= n; i++) 
        S += i * i - (i - 1) * (i - 1);
      
    return S;
}
  
// Driver Code
int main()
{
    int n = 100;
    cout << "The sum of n term is: "
        << summingSeries(n) << endl;
    return 0;
}

Java

// JAVA program to find summation of series
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
  
class GFG 
{
  
    // fuction to calulate sum of series
    static int summingSeries(long n)
    {
        // use of loop to calculate
        // sum of each term
        int S = 0
        for (i = 1; i <= n; i++) 
            S += i * i - (i - 1) * (i - 1);     
          
        return S;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int n = 100;
        System.out.println("The sum of n term is: "
                            summingSeries(n));
    }
}

/div>

Python3

# Python3 program to find summation
# of series
  
def summingSeries(n):
  
    # use of loop to calculate
    # sum of each term
    S = 0
    for i in range(1, n+1): 
        S += i * i - (i - 1) * (i - 1)
      
    return S
  
# Driver Code
n = 100
print("The sum of n term is: "
           summingSeries(n), sep = "")
# This code is contributed by Smitha.

C#

   
// C# program to illustrate...
// Summation of series
using System;
  
class GFG 
{
  
    // function to calculate sum of series
    static int summingSeries(long n)
    {
  
        // Using the pow function calculate
        // the sum of the series
        return (int)Math.Pow(n, 2);
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int n = 100;
        Console.Write("The sum of n term is: "
                        summingSeries(n));
    }
}
  
// This code contribute by Parashar...

PHP

<?php
// PHP program to find 
// summation of series
  
function summingSeries( $n)
{
      
    // use of loop to calculate
    // sum of each term
    $S = 0; 
    for ($i = 1; $i <= $n; $i++) 
         $S += $i * $i - ($i - 1) * 
                       ($i - 1);
      
    return $S;
}
  
// Driver Code
$n = 100;
echo "The sum of n term is: ",
summingSeries($n) ;
  
// This code contribute by vt_m.
?>


Output:



The sum of n term is: 10000

Time complexity – O(N)
Space complexity – O(1)

Efficient Approach
Use of mathematical approach can solve this problem in more efficient way.

Tn = n2 – (n-1)2

Sum of the series is given by (S) = SUM( Tn )

LET US TAKE A EXAMPLE IF
N = 4
It means there should be 4 terms in the series so

1st term = 12 – ( 1 – 1 )2
2nd term = 22 – ( 2 – 1 )2
3th term = 32 – ( 3 – 1 )2
4th term = 42 – ( 3 – 1 )2

SO SUM IS GIVEN BY = (1 – 0) + (4 – 1) + (9 – 4) + (16 – 9)
= 16

FROM THIS WE HAVE NOTICE THAT 1, 4, 9 GET CANCELLED FROM THE SERIES
ONLY 16 IS LEFT WHICH IS EQUAL TO THE SQUARE OF N

So from the above series we notice that each term gets canceled from the next term, only the last term is left which is equal to N2.

C++

// CPP program to illustrate...
// Summation of series
  
#include <bits/stdc++.h>
using namespace std;
  
int summingSeries(long n)
{
    // Sum of n terms is 2^n
    return pow(n, 2);
}
  
// Driver Code
int main()
{
    int n = 100;
    cout << "The sum of n term is: "
         << summingSeries(n) << endl;
    return 0;
}

Java

// JAVA program to illustrate...
// Summation of series
  
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
  
class GFG 
{
  
    // function to calculate sum of series
    static int summingSeries(long n)
    {
  
        // Using the pow function calculate
        // the sum of the series
        return (int)Math.pow(n, 2);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int n = 100;
        System.out.println("The sum of n term is: "
                            summingSeries(n));
    }
}

Python3

# Python3 program to illustrate...
# Summation of series
import math
  
def summingSeries(n):
  
    # Sum of n terms is 2^n
    return math.pow(n, 2)
  
# Driver Code
n = 100
print ("The sum of n term is: "
        summingSeries(n))
# This code is contributed by mits.

C#

// C# program to illustrate...
// Summation of series
using System;
  
class GFG 
{
    // function to calculate sum of series
    static int summingSeries(long n)
    {
        // Using the pow function calculate
        // the sum of the series
        return (int)Math.Pow(n, 2);
    }
  
    // Driver code
    public static void Main()
    {
        int n = 100;
           Console.Write("The sum of n term is: "
                              summingSeries(n));
    }
}
  
// This code is contributed by nitin mittal.

PHP

<?php
// PHP program to illustrate...
// Summation of series
  
function summingSeries($n)
{
    // Sum of n terms is 2^n
    return pow($n, 2);
}
  
// Driver Code
$n = 100;
echo "The sum of n term is: ",
summingSeries($n);
  
// This code contribute by vt_m.
?>


Output:

The sum of n term is: 10000


Time complexity – O(1)
Space complexity – O(1)



This article is attributed to GeeksforGeeks.org

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