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# Difference between sum of the squares of first n natural numbers and square of sum

• Difficulty Level : Easy
• Last Updated : 31 Mar, 2021

Given an integer n, find the absolute difference between sum of the squares of first n natural numbers and square of sum of first n natural numbers.
Examples :

```Input : n = 3
Output : 22.0
Sum of first three numbers is 3 + 2 + 1 = 6
Square of the sum =  36
Sum of squares of first three is 9 + 4 + 1 = 14
Absolute difference = 36 - 14 = 22

Input : n = 10
Output : 2640.0```

Approach :
1. Find the sum of square of first n natural numbers.
2. Find the sum of first n numbers and square it.
3. Find the absolute difference between both the sums and print it.
Below is the implementation of above approach :

## C++

 `// C++ program to find the difference``// between sum of the squares of the``// first n natural numbers and square``// of sum of first n natural number``#include ``using` `namespace` `std;` `int` `Square_Diff(``int` `n){` `int` `l, k, m;` `    ``// Sum of the squares of the``    ``// first n natural numbers is``    ``l = (n * (n + 1) * (2 * n + 1)) / 6;``    ` `    ``// Sum of n naturals numbers``    ``k = (n * (n + 1)) / 2;` `    ``// Square of k``    ``k = k * k;``    ` `    ``// Differences between l and k``    ``m = ``abs``(l - k);``    ` `    ``return` `m;` `}` `// Driver Code``int` `main()``{``    ``int` `n = 10;``    ``cout << Square_Diff(n);``    ``return` `0;``    ` `}` `// This code is contributed by 'Gitanjali' .`

## Java

 `// Java program to find the difference``// between sum of the squares of the``// first n natural numbers and square``// of sum of first n natural number` `public` `class` `GfG{` `static` `int` `Square_Diff(``int` `n){` `int` `l, k, m;``    ``// Sum of the squares of the``    ``// first n natural numbers is``    ``l = (n * (n + ``1``) * (``2` `* n + ``1``)) / ``6``;``    ` `    ``// Sum of n naturals numbers``    ``k = (n * (n + ``1``)) / ``2``;` `    ``// Square of k``    ``k = k * k;``    ` `    ``// Differences between l and k``    ``m = Math.abs(l - k);``    ` `    ``return` `m;` `}` `// Driver Code``public` `static` `void` `main(String s[])``{``    ``int` `n = ``10``;``    ``System.out.println(Square_Diff(n));    ``    ` `}``}``// This code is contributed by 'Gitanjali'.`

## Python

 `# Python3 program to find the difference``# between sum of the squares of the``# first n natural numbers and square``# of sum of first n natural number` `def` `Square_Diff(n):` `    ``# sum of the squares of the``    ``# first n natural numbers is``    ``l ``=` `(n ``*` `(n ``+` `1``) ``*` `(``2` `*` `n ``+` `1``)) ``/` `6``    ` `    ``# sum of n naturals numbers``    ``k ``=` `(n ``*` `(n ``+` `1``)) ``/` `2` `    ``# square of k``    ``k ``=` `k ``*``*` `2``    ` `    ``# Differences between l and k``    ``m ``=` `abs``(l ``-` `k)``    ` `    ``return` `m` `# Driver code``print``(Square_Diff(``10``))`

C#]

``````

<?php
// PHP program to find the difference
// between sum of the squares of the
// first n natural numbers and square
// of sum of first n natural number

function Square_Diff(\$n)
{

\$l;
\$k;
\$m;

// Sum of the squares of the
// first n natural numbers is
\$l = (\$n * (\$n + 1) *
(2 * \$n + 1)) / 6;

// Sum of n naturals numbers
\$k = (\$n * (\$n + 1)) / 2;

// Square of k
\$k = \$k * \$k;

// Differences between
// l and k
\$m = abs(\$l - \$k);

return \$m;

}

// Driver Code
\$n = 10;
echo Square_Diff(\$n);

// This code is contributed by anuj_67 .
?>

``````

## Javascript

 ``

Output :

`2640`

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