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 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’.