Given an integer N, the task is to calculate the sum of all i from 1 to N such that (2i + 1) % 3 = 0.
Examples:
Input: N = 3
Output: 4
For i = 1, 21 + 1 = 3 is divisible by 3.
For i = 2, 22 + 1 = 5 which is not divisible by 3.
For i = 3, 23 + 1 = 9 is divisible by 3.
Hence, sum = 1 + 3 = 4 (for i = 1, 3)Input: N = 13
Output: 49
Approach: If we observe carefully then i will always be an odd number i.e. 1, 3, 5, 7, …... We will use the formula for the sum of the first n odd numbers which is n * n.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the required sum int sumN( int n)
{ // Total odd numbers from 1 to n
n = (n + 1) / 2;
// Sum of first n odd numbers
return (n * n);
} // Driver code int main()
{ int n = 3;
cout << sumN(n);
return 0;
} |
// Java implementation of the approach class GFG {
// Function to return the required sum
static int sum( int n)
{
// Total odd numbers from 1 to n
n = (n + 1 ) / 2 ;
// Sum of first n odd numbers
return (n * n);
}
// Driver code
public static void main(String args[])
{
int n = 3 ;
System.out.println(sum(n));
}
} |
# Python3 implementation of the approach # Function to return the required sum def sumN(n):
# Total odd numbers from 1 to n
n = (n + 1 ) / / 2 ;
# Sum of first n odd numbers
return (n * n);
# Driver code n = 3 ;
print (sumN(n));
# This code is contributed by mits |
// C# implementation of the approach using System;
public class GFG {
// Function to return the required sum
public static int sum( int n)
{
// Total odd numbers from 1 to n
n = (n + 1) / 2;
// Sum of first n odd numbers
return (n * n);
}
// Driver code
public static void Main( string [] args)
{
int n = 3;
Console.WriteLine(sum(n));
}
} // This code is contributed by Shrikant13 |
<?php // PHP implementation of the approach // Function to return the required sum function sumN( $n )
{ // Total odd numbers from 1 to n
$n = (int)(( $n + 1) / 2);
// Sum of first n odd numbers
return ( $n * $n );
} // Driver code $n = 3;
echo sumN( $n );
// This code is contributed by mits ?> |
<script> // Javascript implementation of the approach // Function to return the required sum function sumN(n)
{ // Total odd numbers from 1 to n
n = parseInt((n + 1) / 2);
// Sum of first n odd numbers
return (n * n);
} // Driver code var n = 3;
document.write(sumN(n)); // This code is contributed by noob2000. </script> |
4
Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.