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 n^th term of the sequence is given by:

T_n = 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)); 
    } 
} 

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.

T_n = n² – (n-1)²

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

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

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

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