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.
// 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.*/ |
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Time Complexity : O(n)
Auxiliary Space: O(1)
Method 2:
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
// 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.*/ |
30
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.
// 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)> |
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Time complexity: O(1) as it is doing constant operations
Auxiliary Space: O(1) as it is using constant space
Please refer complete article on Sum of squares of first n natural numbers for more details!