Given a number n, find sum of square of first n odd natural numbers.
Examples :
Input : 3 Output : 35 12 + 32 + 52 = 35 Input : 8 Output : 680 12 + 32 + 52 + 72 + 92 + 112 + 132 + 152
A simple solution is to traverse through n odd numbers and find the sum of square.
Below is the implementation of the approach.
Try It Yourself
// Simple C++ method to find sum of square of
// first n odd numbers.
#include <iostream>
using namespace std;
int squareSum(int n)
{
int sum = 0;
for (int i = 1; i <= n; i++)
sum += (2*i - 1) * (2*i - 1);
return sum;
}
int main()
{
cout << squareSum(8);
return 0;
}
// Simple Java method to
// find sum of square of
// first n odd numbers.
import java.io.*;
class GFG {
static int squareSum(int n)
{
int sum = 0;
for (int i = 1; i <= n; i++)
sum += (2*i - 1) * (2*i - 1);
return sum;
}
//Driver Code
public static void main(String args[])
{
System.out.println(squareSum(8));
}
}
// This code is contributed by
// Nikita tiwari.
# Simple Python method
# to find sum of square
# of first n odd numbers.
def squareSum(n):
sm = 0
for i in range(1, n + 1):
sm += (2 * i - 1) * (2 * i - 1)
return sm
# Driver Code
n=8
print(squareSum(n))
# This code is contributed by Ansu Kumari
// Simple C# method to find
// sum of square of first
// n odd numbers.
using System;
class GFG {
static int squareSum(int n)
{
int sum = 0;
for (int i = 1; i <= n; i++)
sum += (2*i - 1) * (2*i - 1);
return sum;
}
// Driver Code
public static void Main()
{
Console.Write(squareSum(8));
}
}
// This code is contributed by
// vt_m.
<?php
// Simple PHP method to find sum
// of square of first n odd numbers.
function squareSum( $n)
{
$sum = 0;
for ($i = 1; $i <= $n; $i++)
$sum += (2*$i - 1) * (2*$i - 1);
return $sum;
}
echo squareSum(8);
// This code is contributed by Vt_m.
?>
<script>
// Simple Javascript method to find
// sum of square of first n odd numbers.
function squareSum(n)
{
let sum = 0;
for(let i = 1; i <= n; i++)
sum += (2 * i - 1) * (2 * i - 1);
return sum;
}
// Driver code
document.write(squareSum(8));
// This code is contributed by souravmahato348
</script>
Output :
680
Time Complexity: O(n), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
An efficient solution is to apply below formula.
sum = n * (4n2 - 1) / 3 How does it work? Please refer sum of squares of even and odd numbers for proof.
// Efficient C++ method to find sum of
// square of first n odd numbers.
#include <iostream>
using namespace std;
int squareSum(int n)
{
return n*(4*n*n - 1)/3;
}
int main()
{
cout << squareSum(8);
return 0;
}
// Efficient Java method
// to find sum of
// square of first n odd numbers.
import java.io.*;
class GFG {
static int squareSum(int n)
{
return n*(4*n*n - 1)/3;
}
public static void main(String args[])
{
System.out.println(squareSum(8));
}
}
// This code is contributed by
// Nikita tiwari.
# Python3 code to find sum
# of square of first n odd numbers
def squareSum( n ):
return int(n * ( 4 * n * n - 1) / 3)
# driver code
ans = squareSum(8)
print (ans)
# This code is contributed by Saloni Gupta
// Efficient C# method to
// find sum of square of
// first n odd numbers.
using System;
class GFG {
static int squareSum(int n)
{
return n * (4 * n * n - 1)/3;
}
// driver code
public static void Main()
{
Console.Write(squareSum(8));
}
}
// This code is contributed by
// Vt_m.
<?php
// Efficient PHP method to find sum of
// square of first n odd numbers.
function squareSum($n)
{
return $n * (4 * $n * $n - 1) / 3;
}
echo squareSum(8);
// This code is contributed by Vt_m.
?>
<script>
// JavaScript program to find sum of
// square of first n odd numbers.
function squareSum(n)
{
return n*(4*n*n - 1)/3;
}
// Driver code
document.write(squareSum(8));
</script>
Output :
680
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.
