Program to print the sum of the given nth term

Last Updated : 29 Oct, 2023

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 term 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 
{
 
    // function to calculate 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...

Javascript

<script>
// Javascript program to find summation of series
 
function summingSeries(n)
{
    // use of loop to calculate
    // sum of each term
    let S = 0; 
    for (let i = 1; i <= n; i++) 
        S += i * i - (i - 1) * (i - 1);
     
    return S;
}
 
// Driver Code
let n = 100;
document.write("The sum of n term is: " + summingSeries(n));
 
// This code is contributed by rishavmahato348.
</script>

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 n^2
    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  n^2
    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.

Javascript

<script>
// Javascript program to illustrate...
// Summation of series
 
function summingSeries(n)
{
    // Sum of n terms is n^2
    return Math.pow(n, 2);
}
 
// Driver Code
let n = 100;
document.write("The sum of n term is: "
    + summingSeries(n) + "<br>");
     
    // This code is contributed by subham348.
</script>

PHP

<?php
// PHP program to illustrate...
// Summation of series
 
function summingSeries($n)
{
    // Sum of n terms is  n^2
    return pow($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)

Approach#3: Using for loop

Take the value of ‘n’ from the user. Initialize the sum variable to 0. Use a for loop to iterate over the range of 1 to 2*n with a step size of 2. In each iteration, add the current value to the sum variable. After the loop, print the sum of the first n terms of the sequence.

Algorithm

1. Start
2. Take the value of ‘n’ from the user and store it in a variable.
3. Initialize the sum variable to 0.
4. Use a for loop to iterate over the range of 1 to 2*n with a step size of 2.
a. Add the current value to the sum variable.
5. After the loop, print the sum of the first n terms of the sequence.
End

C++

#include <iostream>
using namespace std;
 
int main() {
    // Given value of n
    const int n = 100;
 
    // Initialize the sum
    int sum = 0;
 
    // Loop through odd numbers from 1 to 2*n with step 2
    for (int i = 1; i < 2 * n; i += 2) {
        // Add the current odd number to the sum
        sum += i;
    }
 
    // Print the sum of the first n terms of the sequence
    cout << "Sum of first " << n << " terms of the sequence is: " << sum <<endl;
 
    return 0;
}

Java

public class SumOfOddNumbers {
    public static void main(String[] args)
    {
        // Given value of n
        final int n = 100;
 
        // Initialize the sum
        int sum = 0;
 
        // Loop through odd numbers from 1 to 2*n with step
        // 2
        for (int i = 1; i < 2 * n; i += 2) {
            // Add the current odd number to the sum
            sum += i;
        }
 
        // Print the sum of the first n terms of the
        // sequence
        System.out.println("Sum of first " + n
                           + " terms of the sequence is: "
                           + sum);
    }
}

Python3

n = 100
sum = 0
for i in range(1, 2*n, 2):
    sum += i
print("Sum of first {} terms of the sequence is: {}".format(n, sum))

C#

using System;
 
class Program {
    static void Main()
    {
        // Given value of n
        const int n = 100;
 
        // Initialize the sum
        int sum = 0;
 
        // Loop through odd numbers from 1 to 2*n with step
        // 2
        for (int i = 1; i < 2 * n; i += 2) {
            // Add the current odd number to the sum
            sum += i;
        }
 
        // Print the sum of the first n terms of the
        // sequence
        Console.WriteLine("Sum of first " + n
                          + " terms of the sequence is: "
                          + sum);
    }
}

Javascript

// Given value of n
const n = 100;
 
// Initialize the sum
let sum = 0;
 
// Loop through odd numbers from 1 to 2*n with step 2
for (let i = 1; i < 2 * n; i += 2) {
    // Add the current odd number to the sum
    sum += i;
}
 
// Print the sum of the first n terms of the sequence
console.log(`Sum of first ${n} terms of the sequence is: ${sum}`);

Output

Sum of first 100 terms of the sequence is: 10000

Time Complexity: O(n) The for loop iterates n times.
Space Complexity: O(1) Only a constant amount of extra space is required to store the sum and the loop variable ‘i’.

Program to find the Nth term of the series 0, 5, 14, 27, 44, ........

Kanishk_Verma

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Mathematical

Program to print the sum of the given nth term

C++

Java

Python3

C#

Javascript

PHP

C++

Java

Python3

C#

Javascript

PHP

Approach#3: Using for loop

Algorithm

C++

Java

Python3

C#

Javascript

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