Find the sum of N terms of the series 1^2, (1^2+2^2), (1^2+2^2+3^2), .....

Given a positive integer, N. Find the sum of the first N term of the series-

1², (1²+2²), (1²+2²+3²),....,till N terms

Examples:

Input: N = 3

Output: 20

Input: N = 1

Output: 1

Approach: The sequence is formed by using the following pattern. For any value N-

Given 1^2, (1^2+2^2), (1^2+2^2+3^2),....,till N terms

\sum_{i=1}^{i=n} i = \frac{n(n+1)}{2} \\ \sum_{i=1}^{i=n} i^{2}= \frac{n(n+1)(2n+1)}{6} \\ \sum_{i=1}^{i=n} i^{3}= \frac{n^{2}(n+1)^{2}}{4} \\ 1^{2}+(1^{2}+2^{2}....)+(1^{2}+2^{2}....+n^{2})\\ = \sum_{i=1}^{i=n} (1^{2}+2^{2}...+i^{2})\\ =\sum_{i=1}^{i=n}\sum_{i=1}^{i=i}i^{2}\\ =\sum_{i=1}^{i=n}  \frac{i(i+1)(2i+1)}{6} \\ =\sum_{i=1}^{i=n}  \frac{2i^{3}+3i^{2}+i}{6} \\ =\frac{1}{3}\sum_{i=1}^{i=n} i^{3}+\frac{1}{2}\sum_{i=1}^{i=n} i^{2}+\frac{1}{6}\sum_{i=1}^{i=n} i\\ \text{use above summation formula}\\ =\frac{1}{3}[\frac{n^{2}(n+1)^{2}}{4}]+\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}]+\frac{1}{6}[\frac{n(n+1)}{2}]\\ =\frac{n(n+1)^{2}(n+2)}{12}

S_N = N * (N+1)² * (N+2) / 12

Below is the implementation of the above approach:

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to return sum of
// N term of the series

int findSum(int N){
 
  return N*(N+1)*(N+1)*(N+2)/12;
  
}

// Driver Code

int main()
{
    int N = 3;
    
    cout << findSum(N);
}

// Java program to implement
// the above approach

class GFG {

    // Function to return sum of
    // N term of the series

    static int findSum(int N) {

        return N * (N + 1) * (N + 1) * (N + 2) / 12;

    }

    // Driver Code

    public static void main(String args[]) {
        int N = 3;

        System.out.print(findSum(N));
    }
}

// This code is contributed by Saurabh Jaiswal

# Python 3 program for the above approach

# Function to return sum of
# N term of the series

def findSum(N):
  
  return N*(N+1)*(N+1)*(N+2)//12


# Driver Code
if __name__ == "__main__":
  
    # Value of N
    N = 3    
    print(findSum(N))

# This code is contributed by Abhishek Thakur.

// C# program to implement
// the above approach
using System;
class GFG
{

// Function to return sum of
// N term of the series

static int findSum(int N){
 
  return N*(N+1)*(N+1)*(N+2)/12;
  
}

// Driver Code

public static void Main()
{
    int N = 3;
    
    Console.Write(findSum(N));
}
}

// This code is contributed by Samim Hossain Mondal.

<script>
// Javascript program to implement
// the above approach

// Function to return sum of
// N term of the series

function findSum(N){
 
  return N*(N+1)*(N+1)*(N+2)/12;
  
}

// Driver Code

let N = 3;
document.write(findSum(N));

// This code is contributed by Samim Hossain Mondal.
</script>

20

Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.