Given a number n, the task is to find the sum of the below series upto n terms:
12 – 22 + 32 – 42 + …..
Examples:
Input: n = 2 Output: -3 Explanation: sum = 12 - 22 = 1 - 4 = -3 Input: n = 3 Output: 6 Explanation: sum = 12 - 22 + 32 = 1 - 4 + 9 = 6
Naive Approach:
This method involves simply running a loop of i from 1 to n and if i is odd then simply add its square to the result it i is even then simply subtract square of it to the result.
Below is the implementation of the above approach:
// C++ program to find sum of series // 1^2 - 2^2 + 3^3 - 4^4 + ... #include <bits/stdc++.h> using namespace std;
// Function to find sum of series int sum_of_series( int n)
{ int result = 0;
for ( int i = 1; i <= n; i++) {
// If i is even
if (i % 2 == 0)
result = result - pow (i, 2);
// If i is odd
else
result = result + pow (i, 2);
}
// return the result
return result;
} // Driver Code int main( void )
{ // Get n
int n = 3;
// Find the sum
cout << sum_of_series(n) << endl;
// Get n
n = 10;
// Find the sum
cout << sum_of_series(n) << endl;
} |
// Java Program to find sum of series // 1^2 - 2^2 + 3^3 - 4^4 + ... import java.util.*;
import java.lang.*;
class GFG
{ // Function to find sum of series static int sum_of_series( int n)
{ int result = 0 ;
for ( int i = 1 ; i <= n; i++)
{
// If i is even
if (i % 2 == 0 )
result = result -
( int )Math.pow(i, 2 );
// If i is odd
else
result = result +
( int )Math.pow(i, 2 );
}
// return the result
return result;
} // Driver Code public static void main(String args[])
{ // Get n
int n = 3 ;
// Find the sum
System.out.println(sum_of_series(n));
// Get n
n = 10 ;
// Find the sum
System.out.println(sum_of_series(n));
} } // This code is contributed // by Akanksha Rai(Abby_akku) |
# Python3 program to find sum of series # 1^2 - 2^2 + 3^3 - 4^4 + ... # Function to find sum of series def sum_of_series(n):
result = 0
for i in range ( 1 , n + 1 ) :
# If i is even
if (i % 2 = = 0 ):
result = result - pow (i, 2 )
# If i is odd
else :
result = result + pow (i, 2 )
# return the result
return result
# Driver Code if __name__ = = "__main__" :
# Get n
n = 3
# Find the sum
print (sum_of_series(n))
# Get n
n = 10
# Find the sum
print (sum_of_series(n))
# This code is contributed # by ChitraNayal |
// C# Program to find sum of series // 1^2 - 2^2 + 3^3 - 4^4 + ... using System;
class GFG
{ // Function to find sum of series static int sum_of_series( int n)
{ int result = 0;
for ( int i = 1; i <= n; i++)
{
// If i is even
if (i % 2 == 0)
result = result -
( int )Math.Pow(i, 2);
// If i is odd
else
result = result +
( int )Math.Pow(i, 2);
}
// return the result
return result;
} // Driver Code public static void Main()
{ // Get n
int n = 3;
// Find the sum
Console.WriteLine(sum_of_series(n));
// Get n
n = 10;
// Find the sum
Console.WriteLine(sum_of_series(n));
} } // This code is contributed // by Akanksha Rai(Abby_akku) |
<?php // PHP program to find sum of series // 1^2 - 2^2 + 3^3 - 4^4 + ... // Function to find sum of series function sum_of_series( $n )
{ $result = 0;
for ( $i = 1; $i <= $n ; $i ++)
{
// If i is even
if ( $i % 2 == 0)
$result = $result - pow( $i , 2);
// If i is odd
else
$result = $result + pow( $i , 2);
}
// return the result
return $result ;
} // Driver Code // Get n $n = 3;
// Find the sum echo sum_of_series( $n ), "\n" ;
// Get n $n = 10;
// Find the sum echo sum_of_series( $n ), "\n" ;
// This Code is Contributed by anuj_67 ?> |
<script> // javascript Program to find sum of series // 1^2 - 2^2 + 3^3 - 4^4 + ... // Function to find sum of series function sum_of_series(n)
{ var result = 0;
for (i = 1; i <= n; i++)
{
// If i is even
if (i % 2 == 0)
result = result -
parseInt(Math.pow(i, 2));
// If i is odd
else
result = result +
parseInt(Math.pow(i, 2));
}
// return the result
return result;
} // Driver Code // Get n var n = 3;
// Find the sum document.write(sum_of_series(n)+ "<br>" );
// Get n n = 10; // Find the sum document.write(sum_of_series(n)); // This code is contributed by 29AjayKumar </script> |
Output:
6 -55
Time Complexity: O(n)
Auxiliary Space: O(1)
Efficient Approach
It is based on condition of n
If n is even:
If n is odd:
Below is the implementation of the above approach:
// C++ Program to find sum of series // 1^2 - 2^2 +3^3 -4^4 + ... #include <bits/stdc++.h> using namespace std;
// Function to find sum of series int sum_of_series( int n)
{ int result = 0;
// If n is even
if (n % 2 == 0) {
result = -(n * (n + 1)) / 2;
}
// If n is odd
else {
result = (n * (n + 1)) / 2;
}
// return the result
return result;
} // Driver Code int main( void )
{ // Get n
int n = 3;
// Find the sum
cout << sum_of_series(n) << endl;
// Get n
n = 10;
// Find the sum
cout << sum_of_series(n) << endl;
} |
// Java Program to find sum of series // 1^2 - 2^2 +3^3 -4^4 + ... import java.util.*;
import java.lang.*;
class GFG
{ // Function to find sum of series static int sum_of_series( int n)
{ int result = 0 ;
// If n is even
if (n % 2 == 0 )
{
result = -(n * (n + 1 )) / 2 ;
}
// If n is odd
else
{
result = (n * (n + 1 )) / 2 ;
}
// return the result
return result;
} // Driver Code public static void main(String args[])
{ // Get n
int n = 3 ;
// Find the sum
System.out.println(sum_of_series(n));
// Get n
n = 10 ;
// Find the sum
System.out.println(sum_of_series(n));
} } // This code is contributed // by Akanksha Rai(Abby_akku) |
# Python3 Program to find sum of series # 1^2 - 2^2 +3^3 -4^4 + ... # Function to find sum of series def sum_of_series(n) :
result = 0
# If n is even
if (n % 2 = = 0 ) :
result = - (n * (n + 1 )) / / 2
# If n is odd
else :
result = (n * (n + 1 )) / / 2
# return the result
return result
# Driver Code if __name__ = = "__main__" :
# Get n
n = 3
# Find the sum
print (sum_of_series(n))
# Get n
n = 10
# Find the sum
print (sum_of_series(n))
# This code is contributed by Ryuga |
// C# Program to find sum of series // 1^2 - 2^2 +3^3 -4^4 + ... using System;
class GFG
{ // Function to find sum of series static int sum_of_series( int n)
{ int result = 0;
// If n is even
if (n % 2 == 0)
{
result = -(n * (n + 1)) / 2;
}
// If n is odd
else
{
result = (n * (n + 1)) / 2;
}
// return the result
return result;
} // Driver Code public static void Main()
{ // Get n
int n = 3;
// Find the sum
Console.WriteLine(sum_of_series(n));
// Get n
n = 10;
// Find the sum
Console.WriteLine(sum_of_series(n));
} } // This code is contributed // by Akanksha Rai(Abby_akku) |
<?php // PHP program to find sum of series // 1^2 - 2^2 +3^3 -4^4 + ... // Function to find sum of series function sum_of_series( $n )
{ $result = 0;
// If n is even
if ( $n % 2 == 0)
{
$result = -( $n * ( $n + 1)) / 2;
}
// If n is odd
else
{
$result = ( $n * ( $n + 1)) / 2;
}
// return the result
return $result ;
} // Driver Code // Get n $n = 3;
// Find the sum echo sum_of_series( $n );
echo ( "\n" );
// Get n $n = 10;
// Find the sum echo sum_of_series( $n );
echo ( "\n" );
// Get n $n = 10;
// This code is contributed // by Shivi_Aggarwal ?> |
<script> // Javascript Program to find sum of series // 1^2 - 2^2 +3^3 -4^4 + ... // Function to find sum of series
function sum_of_series( n) {
let result = 0;
// If n is even
if (n % 2 == 0) {
result = -(n * (n + 1)) / 2;
}
// If n is odd
else {
result = (n * (n + 1)) / 2;
}
// return the result
return result;
}
// Driver Code
// Get n
let n = 3;
// Find the sum
document.write(sum_of_series(n)+ "<br/>" );
// Get n
n = 10;
// Find the sum
document.write(sum_of_series(n));
// This code is contributed by 29AjayKumar </script> |
Output:
6 -55
Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.