Given a positive integer N. The task is to find 12 + 22 + 32 + ….. + N2.
Examples :
Input : N = 4 Output : 30 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30 Input : N = 5 Output : 55
Method 1: O(N) The idea is to run a loop from 1 to n and for each i, 1 <= i <= n, find i2 to sum.
Below is the implementation of this approach
// CPP Program to find sum of square of first n natural numbers #include <bits/stdc++.h> using namespace std;
// Return the sum of square of first n natural numbers int squaresum( int n)
{ // Iterate i from 1 and n
// finding square of i and add to sum.
int sum = 0;
for ( int i = 1; i <= n; i++)
sum += (i * i);
return sum;
} // Driven Program int main()
{ int n = 4;
cout << squaresum(n) << endl;
return 0;
} |
// Java Program to find sum of // square of first n natural numbers import java.io.*;
class GFG {
// Return the sum of square of first n natural numbers
static int squaresum( int n)
{
// Iterate i from 1 and n
// finding square of i and add to sum.
int sum = 0 ;
for ( int i = 1 ; i <= n; i++)
sum += (i * i);
return sum;
}
// Driven Program
public static void main(String args[]) throws IOException
{
int n = 4 ;
System.out.println(squaresum(n));
}
} /*This code is contributed by Nikita Tiwari.*/ |
# Python3 Program to # find sum of square # of first n natural # numbers # Return the sum of # square of first n # natural numbers def squaresum(n) :
# Iterate i from 1
# and n finding
# square of i and
# add to sum.
sm = 0
for i in range ( 1 , n + 1 ) :
sm = sm + (i * i)
return sm
# Driven Program n = 4
print (squaresum(n))
# This code is contributed by Nikita Tiwari.*/ |
// C# Program to find sum of // square of first n natural numbers using System;
class GFG {
// Return the sum of square of first
// n natural numbers
static int squaresum( int n)
{
// Iterate i from 1 and n
// finding square of i and add to sum.
int sum = 0;
for ( int i = 1; i <= n; i++)
sum += (i * i);
return sum;
}
// Driven Program
public static void Main()
{
int n = 4;
Console.WriteLine(squaresum(n));
}
} /* This code is contributed by vt_m.*/ |
<?php // PHP Program to find sum of // square of first n natural numbers // Return the sum of square of // first n natural numbers function squaresum( $n )
{ // Iterate i from 1 and n
// finding square of i and
// add to sum.
$sum = 0;
for ( $i = 1; $i <= $n ; $i ++)
$sum += ( $i * $i );
return $sum ;
} // Driven Code $n = 4;
echo (squaresum( $n ));
// This code is contributed by Ajit. ?> |
<script> // Javascript Program to find sum of square of first n natural numbers // Return the sum of square of first n natural numbers function squaresum(n)
{ // Iterate i from 1 and n
// finding square of i and add to sum.
let sum = 0;
for (let i = 1; i <= n; i++)
sum += (i * i);
return sum;
} // Driven Program let n = 4;
document.write(squaresum(n) + "<br>" );
// This code is contributed by Mayank Tyagi </script> |
Output :
30
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 2: O(1)
Sum of squares of first N natural numbers = (N*(N+1)*(2*N+1))/6
For example
For N=4, Sum = ( 4 * ( 4 + 1 ) * ( 2 * 4 + 1 ) ) / 6
= 180 / 6
= 30
For N=5, Sum = ( 5 * ( 5 + 1 ) * ( 2 * 5 + 1 ) ) / 6
= 55
Proof:
We know, (k + 1)3 = k3 + 3 * k2 + 3 * k + 1 We can write the above identity for k from 1 to n: 23 = 13 + 3 * 12 + 3 * 1 + 1 ......... (1) 33 = 23 + 3 * 22 + 3 * 2 + 1 ......... (2) 43 = 33 + 3 * 32 + 3 * 3 + 1 ......... (3) 53 = 43 + 3 * 42 + 3 * 4 + 1 ......... (4) ... n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ......... (n - 1) (n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ......... (n) Putting equation (n - 1) in equation n, (n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1 = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1 By putting all equation, we get (n + 1)3 = 13 + 3 * ? k2 + 3 * ? k + ? 1 n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * ? k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 3 * n = 3 * ? k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * ? k2 n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * ? k2 n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * ? k2 n * (n + 1) * (n + 2 - 3/2) = 3 * ? k2 n * (n + 1) * (2 * n + 1)/2 = 3 * ? k2 n * (n + 1) * (2 * n + 1)/6 = ? k2
Below is the implementation of this approach:
// CPP Program to find sum // of square of first n // natural numbers #include <bits/stdc++.h> using namespace std;
// Return the sum of square of // first n natural numbers int squaresum( int n)
{ return (n * (n + 1) * (2 * n + 1)) / 6;
} // Driven Program int main()
{ int n = 4;
cout << squaresum(n) << endl;
return 0;
} |
// Java Program to find sum // of square of first n // natural numbers import java.io.*;
class GFG {
// Return the sum of square
// of first n natural numbers
static int squaresum( int n)
{
return (n * (n + 1 ) * ( 2 * n + 1 )) / 6 ;
}
// Driven Program
public static void main(String args[])
throws IOException
{
int n = 4 ;
System.out.println(squaresum(n));
}
} /*This code is contributed by Nikita Tiwari.*/ |
# Python3 Program to # find sum of square # of first n natural # numbers # Return the sum of # square of first n # natural numbers def squaresum(n) :
return (n * (n + 1 ) * ( 2 * n + 1 )) / / 6
# Driven Program n = 4
print (squaresum(n))
#This code is contributed by Nikita Tiwari. |
// C# Program to find sum // of square of first n // natural numbers using System;
class GFG {
// Return the sum of square
// of first n natural numbers
static int squaresum( int n)
{
return (n * (n + 1) * (2 * n + 1)) / 6;
}
// Driven Program
public static void Main()
{
int n = 4;
Console.WriteLine(squaresum(n));
}
} /*This code is contributed by vt_m.*/ |
<?php // PHP Program to find sum // of square of first n // natural numbers // Return the sum of square of // first n natural numbers function squaresum( $n )
{ return ( $n * ( $n + 1) *
(2 * $n + 1)) / 6;
} // Driven Code $n = 4;
echo (squaresum( $n ));
// This code is contributed by Ajit. ?> |
<script> // Javascript program to find sum // of square of first n // natural numbers // Return the sum of square of // first n natural numbers function squaresum(n)
{ return parseInt((n * (n + 1) *
(2 * n + 1)) / 6);
} // Driver code let n = 4; document.write(squaresum(n)); // This code is contributed by rishavmahato348 </script> |
Output :
30
Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken
Avoiding early overflow:
For large n, the value of (n * (n + 1) * (2 * n + 1)) would overflow. We can avoid overflow up to some extent using the fact that n*(n+1) must be divisible by 2.
// CPP Program to find sum of square of first // n natural numbers. This program avoids // overflow upto some extent for large value // of n. #include <bits/stdc++.h> using namespace std;
// Return the sum of square of first n natural // numbers int squaresum( int n)
{ return (n * (n + 1) / 2) * (2 * n + 1) / 3;
} // Driven Program int main()
{ int n = 4;
cout << squaresum(n) << endl;
return 0;
} |
# Python Program to find sum of square of first # n natural numbers. This program avoids # overflow upto some extent for large value # of n.y def squaresum(n):
return (n * (n + 1 ) / 2 ) * ( 2 * n + 1 ) / 3
# main() n = 4
print (squaresum(n));
# Code Contributed by Mohit Gupta_OMG <(0_o)> |
// Java Program to find sum of square of first // n natural numbers. This program avoids // overflow upto some extent for large value // of n. import java.io.*;
import java.util.*;
class GFG
{ // Return the sum of square of first n natural
// numbers
public static int squaresum( int n)
{ return (n * (n + 1 ) / 2 ) * ( 2 * n + 1 ) / 3 ;
} public static void main (String[] args)
{
int n = 4 ;
System.out.println(squaresum(n));
}
} // Code Contributed by Mohit Gupta_OMG <(0_o)> |
// C# Program to find sum of square of first // n natural numbers. This program avoids // overflow upto some extent for large value // of n. using System;
class GFG {
// Return the sum of square of
// first n natural numbers
public static int squaresum( int n)
{
return (n * (n + 1) / 2) * (2 * n + 1) / 3;
}
// Driver Code
public static void Main()
{
int n = 4;
Console.WriteLine(squaresum(n));
}
} // This Code is Contributed by vt_m.> |
<?php // PHP Program to find // sum of square of first // n natural numbers. // This program avoids // overflow upto some // extent for large value // of n. // Return the sum of square // of first n natural numbers function squaresum( $n )
{ return ( $n * ( $n + 1) / 2) *
(2 * $n + 1) / 3;
} // Driver Code
$n = 4;
echo squaresum( $n ) ;
// This code is contributed by vt_m. ?> |
<script> // javascript Program to find sum of square of first // n natural numbers. This program avoids // overflow upto some extent for large value // of n. // Return the sum of square of first n natural // numbers function squaresum( n)
{ return (n * (n + 1) / 2) * (2 * n + 1) / 3;
} // Driven Program let n = 4;
document.write(squaresum(n));
// This code contributed by aashish1995 </script> |
Output:
30
Time complexity: O(1) since performing constant operations
Space complexity: O(1) since using constant variables