Find sum of the series 1+22+333+4444+…… upto n terms

Given a number N. The task is to find the sum of the below series up to N-th term:

1 + 22 + 333 + 4444 + …up to n terms

Examples: 

Input: N = 3
Output: 356

Input: N = 10 
Output: 12208504795

Approach:

Below is the implementation of the above approach:

C++

// CPP program to find the sum
// of given series

#include <iostream>
#include <math.h>

using namespace std;

// Function to calculate sum
int findSum(int n)
{
    // Return sum
    return (pow(10, n + 1) * (9 * n - 1) + 10) /
                    pow(9, 3) - n * (n + 1) / 18;
}

// Driver code
int main()
{
    int n = 3;

    cout << findSum(n);

    return 0;
}

Java

// Java Program to find
// Sum of first n terms
import java.util.*;

class solution
{
static int calculateSum(int n)
{

// Returning the final sum
return ((int)Math.pow(10, n + 1) * (9 * n - 1) + 10) /
                (int)Math.pow(9, 3) - n * (n + 1) / 18;
}

// Driver code
public static void main(String ar[])
{
// no. of terms to find the sum
int n=3;
System.out.println("Sum= "+ calculateSum(n));

}
}

//This code is contributed by Surendra_Gangwar

Python 3

# Python program to find the sum of given series.


# Function to calculate sum
def solve_sum(n):
    # Return sum
    return (pow(10, n + 1)*(9 * n - 1)+10)/pow(9, 3)-n*(n + 1)/18

# driver code
n = 3

print(int(solve_sum(n)))

C#

// C# Program to find
// Sum of first n terms
using System;
class solution
{
static int calculateSum(int n)
{

// Returning the final sum
return ((int)Math.Pow(10, n + 1) * (9 * n - 1) + 10) /
                (int)Math.Pow(9, 3) - n * (n + 1) / 18;
}

// Driver code
public static void  Main()
{
// no. of terms to find the sum
int n=3;
Console.WriteLine("Sum= "+ calculateSum(n));

}
}

//This code is contributed by inder_verma.

PHP

<?php
// PHP program to find the sum
// of given series

// Function to calculate sum
function findSum($n)
{
    // Return sum
    return (pow(10, $n + 1) *
               (9 * $n - 1) + 10) /
            pow(9, 3) - $n * ($n + 1) / 18;
}

// Driver code

$n = 3;

echo findSum($n);

// This code is contributed
// by inder_verma.
?>

Javascript

<script>
// Javascript Program to find
// Sum of first n terms

    function calculateSum( n)
    {

        // Returning the const sum
        return (parseInt(Math.pow(10, n + 1)) * (9 * n - 1) + 10) /
                parseInt(Math.pow(9, 3)) - n * (n + 1) / 18;
     }

    // Driver code

    // no. of terms to find the sum
    let n = 3;
    document.write("Sum= " + calculateSum(n));

// This code is contributed by 29AjayKumar
</script>

Output:

356

Time Complexity: O(logn), where n represents the given integer, as we have used pow function.
Auxiliary Space: O(1), no extra space is required, so it is a constant.