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# Program to find the sum of a Series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + … + (n*n)

• Difficulty Level : Medium
• Last Updated : 20 Feb, 2023

You have been given a series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + … + (n*n), find out the sum of the series till nth term.
Examples :

```Input : n = 3
Output : 14
Explanation : (1*1) + (2*2) + (3*3)

Input : n = 5
Output : 55
Explanation : (1*1) + (2*2) + (3*3) + (4*4) + (5*5)```

## C++

 `// CPP program to calculate the following series ` `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate the following series ` `int` `Series(``int` `n) ` `{ ` `    ``int` `i; ` `    ``int` `sums = 0; ` `    ``for` `(i = 1; i <= n; i++) ` `        ``sums += (i * i); ` `    ``return` `sums; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``int` `res = Series(n); ` `    ``cout<

## C

 `// C program to calculate the following series ` `#include ` ` `  `// Function to calculate the following series ` `int` `Series(``int` `n) ` `{ ` `    ``int` `i; ` `    ``int` `sums = 0; ` `    ``for` `(i = 1; i <= n; i++) ` `        ``sums += (i * i); ` `    ``return` `sums; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``int` `res = Series(n); ` `    ``printf``(``"%d"``, res); ` `} `

## Java

 `// Java program to calculate the following series ` `import` `java.io.*; ` `class` `GFG { ` ` `  `    ``// Function to calculate the following series ` `    ``static` `int` `Series(``int` `n) ` `    ``{ ` `        ``int` `i; ` `        ``int` `sums = ``0``; ` `        ``for` `(i = ``1``; i <= n; i++) ` `            ``sums += (i * i); ` `        ``return` `sums; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``3``; ` `        ``int` `res = Series(n); ` `        ``System.out.println(res); ` `    ``} ` `} `

## Python

 `# Python program to calculate the following series ` `def` `Series(n): ` `    ``sums ``=` `0` `    ``for` `i ``in` `range``(``1``, n ``+` `1``): ` `        ``sums ``+``=` `(i ``*` `i); ` `    ``return` `sums ` ` `  `# Driver Code ` `n ``=` `3` `res ``=` `Series(n) ` `print``(res) `

## C#

 `// C# program to calculate the following series ` `using` `System; ` `class` `GFG { ` ` `  `    ``// Function to calculate the following series ` `    ``static` `int` `Series(``int` `n) ` `    ``{ ` `        ``int` `i; ` `        ``int` `sums = 0; ` `        ``for` `(i = 1; i <= n; i++) ` `            ``sums += (i * i); ` `        ``return` `sums; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 3; ` `        ``int` `res = Series(n); ` `        ``Console.Write(res); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

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## Javascript

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Output

`14`

Time Complexity: O(n)

Auxiliary Space: O(1)

Efficient Approach:

As we know the sum of squares of the first N natural numbers = (N*(N+1)*(2*N+1))/6.

Below is the implementation of this approach:

## C++

 `// CPP program to calculate the following series ` `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate the following series ` `int` `Series(``int` `n) ` `{ ` `    ``return` `(n * (n + 1) * (2 * n + 1)) / 6; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``int` `res = Series(n); ` `    ``cout<

## Java

 `// Java program to calculate the following series ` `import` `java.io.*; ` `class` `GFG { ` ` `  `  ``// Function to calculate the following series ` `  ``static` `int` `Series(``int` `n) ` `  ``{ ` `    ``return` `(n * (n + ``1``) * (``2` `* n + ``1``)) / ``6``; ` `  ``} ` ` `  `  ``// Driver function ` `  ``public` `static` `void` `main (String[] args) { ` `    ``int` `n = ``3``; ` `    ``int` `res = Series(n); ` `    ``System.out.println(res); ` ` `  `  ``} ` `} ` ` `  `// This code is contributed by Aman Kumar. `

## Python3

 `# Python program to calculate the following series ` ` `  `# Function to calculate the following series ` `def` `Series(n): ` `    ``return` `(n ``*` `(n``+``1``) ``*` `(``2` `*` `n ``+` `1``)) ``/` `6` ` `  `n ``=` `3` `res ``=` `Series(n) ` `print``(``int``(res)) ` ` `  `# This code is contributed by lokeshmvs21. `

## C#

 `using` `System; ` ` `  `public` `class` `GFG { ` `  ``static` `int` `Series(``int` `n) ` `  ``{ ` `    ``return` `(n * (n + 1) * (2 * n + 1)) / 6; ` `  ``} ` ` `  `  ``static` `public` `void` `Main() ` `  ``{ ` ` `  `    ``int` `n = 3; ` `    ``Console.Write(Series(n)); ` `  ``} ` `} ` ` `  `// This code is contributed by garg28harsh.`

## Javascript

 `// Javascript program to calculate the following series ` ` `  `    ``// Function to calculate the following series ` `    ``function` `Series(n) ` `{ ` `    ``return` `(n * (n + 1) * (2 * n + 1)) / 6; ` `} ` ` `  `// Driver Code ` `let n = 3; ` `console.log(Series(n)); ` ` `  `// This code is contributed by garg28harsh.`

Output

`14`

Time Complexity: O(1)

Auxiliary Space: O(1)
Please refer below post for O(1) solution.
Sum of squares of first n natural numbers

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