# Find the sum of numbers from 1 to n excluding those which are powers of K

Given two integer N and K, the task is to find the sum of all the numbers from the range [1, N] excluding those which are powers of K.

Examples:

Input: N = 10, K = 3
Output: 42
2 + 4 + 5 + 6 + 7 + 8 + 10 = 42
1, 3 and 9 are excluded as they are powers of 3.

Input: N = 200, K = 30
Output: 20069

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Find the sum of the following series:

1. pwrK: The sum of all the powers of K from [1, N] i.e. K0 + K1 + K2 + … + Kr such that Kr ≤ N
2. sumAll: The sum of all the integers from the range [1, N] i.e. (N * (N + 1)) / 2.

The result will be sumAll – pwrK

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define ll long long int ` ` `  `// Function to return the sum of all the ` `// powers of k from the range [1, n] ` `ll sumPowersK(ll n, ll k) ` `{ ` ` `  `    ``// To store the sum of the series ` `    ``ll sum = 0, num = 1; ` ` `  `    ``// While current power of k <= n ` `    ``while` `(num <= n) { ` ` `  `        ``// Add current power to the sum ` `        ``sum += num; ` ` `  `        ``// Next power of k ` `        ``num *= k; ` `    ``} ` ` `  `    ``// Return the sum of the series ` `    ``return` `sum; ` `} ` ` `  `// Find to return the sum of the ` `// elements from the range [1, n] ` `// excluding those which are powers of k ` `ll getSum(ll n, ll k) ` `{ ` `    ``// Sum of all the powers of k from [1, n] ` `    ``ll pwrK = sumPowersK(n, k); ` ` `  `    ``// Sum of all the elements from [1, n] ` `    ``ll sumAll = (n * (n + 1)) / 2; ` ` `  `    ``// Return the required sum ` `    ``return` `(sumAll - pwrK); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``ll n = 10, k = 3; ` ` `  `    ``cout << getSum(n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the sum of all the ` `// powers of k from the range [1, n] ` `static` `long` `sumPowersK(``long` `n, ``long` `k) ` `{ ` ` `  `    ``// To store the sum of the series ` `    ``long` `sum = ``0``, num = ``1``; ` ` `  `    ``// While current power of k <= n ` `    ``while` `(num <= n)  ` `    ``{ ` ` `  `        ``// Add current power to the sum ` `        ``sum += num; ` ` `  `        ``// Next power of k ` `        ``num *= k; ` `    ``} ` ` `  `    ``// Return the sum of the series ` `    ``return` `sum; ` `} ` ` `  `// Find to return the sum of the ` `// elements from the range [1, n] ` `// excluding those which are powers of k ` `static` `long` `getSum(``long` `n, ``long` `k) ` `{ ` `    ``// Sum of all the powers of k from [1, n] ` `    ``long` `pwrK = sumPowersK(n, k); ` ` `  `    ``// Sum of all the elements from [1, n] ` `    ``long` `sumAll = (n * (n + ``1``)) / ``2``; ` ` `  `    ``// Return the required sum ` `    ``return` `(sumAll - pwrK); ` `} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``long` `n = ``10``, k = ``3``; ` `        ``System.out.println( getSum(n, k)); ` ` `  `    ``} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the sum of all the  ` `# powers of k from the range [1, n]  ` `def` `sumPowersK(n, k) : ` ` `  `    ``# To store the sum of the series  ` `    ``sum` `=` `0``; num ``=` `1``;  ` ` `  `    ``# While current power of k <= n  ` `    ``while` `(num <``=` `n) : ` ` `  `        ``# Add current power to the sum  ` `        ``sum` `+``=` `num;  ` ` `  `        ``# Next power of k  ` `        ``num ``*``=` `k;  ` ` `  `    ``# Return the sum of the series  ` `    ``return` `sum``;  ` `     `  ` `  `# Find to return the sum of the  ` `# elements from the range [1, n]  ` `# excluding those which are powers of k  ` `def` `getSum(n, k) : ` ` `  `    ``# Sum of all the powers of k from [1, n]  ` `    ``pwrK ``=` `sumPowersK(n, k);  ` ` `  `    ``# Sum of all the elements from [1, n]  ` `    ``sumAll ``=` `(n ``*` `(n ``+` `1``)) ``/` `2``;  ` ` `  `    ``# Return the required sum  ` `    ``return` `(sumAll ``-` `pwrK);  ` ` `  ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``n ``=` `10``; k ``=` `3``;  ` ` `  `    ``print``(getSum(n, k));  ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the sum of all the ` `// powers of k from the range [1, n] ` `static` `long` `sumPowersK(``long` `n, ``long` `k) ` `{ ` ` `  `    ``// To store the sum of the series ` `    ``long` `sum = 0, num = 1; ` ` `  `    ``// While current power of k <= n ` `    ``while` `(num <= n)  ` `    ``{ ` ` `  `        ``// Add current power to the sum ` `        ``sum += num; ` ` `  `        ``// Next power of k ` `        ``num *= k; ` `    ``} ` ` `  `    ``// Return the sum of the series ` `    ``return` `sum; ` `} ` ` `  `// Find to return the sum of the ` `// elements from the range [1, n] ` `// excluding those which are powers of k ` `static` `long` `getSum(``long` `n, ``long` `k) ` `{ ` `    ``// Sum of all the powers of k from [1, n] ` `    ``long` `pwrK = sumPowersK(n, k); ` ` `  `    ``// Sum of all the elements from [1, n] ` `    ``long` `sumAll = (n * (n + 1)) / 2; ` ` `  `    ``// Return the required sum ` `    ``return` `(sumAll - pwrK); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main ()  ` `{ ` `    ``long` `n = 10, k = 3; ` `    ``Console.WriteLine( getSum(n, k)); ` ` `  `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

Output:

```42
``` My Personal Notes arrow_drop_up Third year Department of Information Technology Jadavpur University

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Improved By : vt_m, AnkitRai01