# Sum of the series Kn + ( K(n-1) * (K-1)1 ) + ( K(n-2) * (K-1)2 ) + ……. (K-1)n

Given a value K and n, the task is to find the sum of the below series:

Kn + ( K(n-1) * (K-1)1 ) + ( K(n-2) * (K-1)2 ) + ……. (K-1)n

Examples:

```Input:  n = 3, K = 3
Output: 65
Explanation:
(3*3*3) + (3*3*2) + (3*2*2) + (2*2*2)
= 27 + 18 + 12 + 8
= 65

Input: n = 4, k = 2
Output: 31
Explanation:
(2*2*2*2) + (2*2*2*1)+ (2*2*1*1) + (2*1*1*1) + (1*1*1*1)
= 16 + 8 + 4 + 2 + 1
= 31

```

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

1. Simple Approach: O(n2)
1. Total number of term in series = n+1
2. Calculate each term separately, and add them:

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return sum ` `int` `sum(``int` `k, ``int` `n) ` `{ ` ` `  `    ``int` `sum = 0; ` `    ``for` `(``int` `i = 0; i <= n; i++) { ` `        ``int` `p = 1; ` ` `  `        ``for` `(``int` `j = 0; j < n - i; j++) { ` `            ``p = p * k; ` `        ``} ` ` `  `        ``for` `(``int` `j = 0; j < i; j++) { ` `            ``p = p * (k - 1); ` `        ``} ` ` `  `        ``sum = sum + p; ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``int` `K = 3; ` `    ``cout << sum(K, n); ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG { ` `    ``// Function to return sum ` `    ``static` `int` `sum(``int` `k, ``int` `n) ` `    ``{ ` ` `  `        ``int` `sum = ``0``; ` `        ``for` `(``int` `i = ``0``; i <= n; i++) { ` `            ``int` `p = ``1``; ` ` `  `            ``for` `(``int` `j = ``0``; j < n - i; j++) { ` `                ``p = p * k; ` `            ``} ` ` `  `            ``for` `(``int` `j = ``0``; j < i; j++) { ` `                ``p = p * (k - ``1``); ` `            ``} ` ` `  `            ``sum = sum + p; ` `        ``} ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``3``; ` `        ``int` `K = ``3``; ` `        ``System.out.println(sum(K, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Code_Mech `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return sum ` `def` `Sum``(k, n): ` ` `  `    ``Summ ``=` `0` `    ``for` `i ``in` `range``(n ``+` `1``): ` `        ``p ``=` `1` ` `  `        ``for` `j ``in` `range``(n ``-` `i): ` `            ``p ``=` `p ``*` `k ` ` `  `        ``for` `j ``in` `range``(i): ` `            ``p ``=` `p ``*` `(k ``-` `1``) ` `         `  `        ``Summ ``=` `Summ ``+` `p ` `     `  `    ``return` `Summ ` ` `  `# Driver code ` `n ``=` `3` `K ``=` `3` `print``(``Sum``(K, n)) ` ` `  `# This code is contributed by mohit kumar `

## C#

 `// C# implementation of the approach ` ` `  `using` `System; ` ` `  `class` `GFG { ` `    ``// Function to return sum ` `    ``static` `int` `sum(``int` `k, ``int` `n) ` `    ``{ ` ` `  `        ``int` `sum = 0; ` `        ``for` `(``int` `i = 0; i <= n; i++) { ` `            ``int` `p = 1; ` ` `  `            ``for` `(``int` `j = 0; j < n - i; j++) { ` `                ``p = p * k; ` `            ``} ` ` `  `            ``for` `(``int` `j = 0; j < i; j++) { ` `                ``p = p * (k - 1); ` `            ``} ` ` `  `            ``sum = sum + p; ` `        ``} ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 3; ` `        ``int` `K = 3; ` `        ``Console.WriteLine(sum(K, n)); ` `    ``} ` `    ``// This code is contributed by Ryuga ` `} `

## PHP

 ` `

Output:

```65
```

Time Complexity: O( n^2 )

2. Second Approach: O(n)

It is observed that, given series is Geometric Progression, with common ratio = (K – 1)/K
So, above expression can be simplified as: Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// Function to return sum ` `int` `sum(``int` `k, ``int` `n) ` `{ ` `    ``int` `sum ` `        ``= ``pow``(k, n + 1) ` `          ``- ``pow``(k - 1, n + 1); ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``int` `K = 3; ` `    ``cout << sum(K, n); ` `} `

## Java

 `// Java implementation of above approach ` `class` `GFG ` `{ ` ` `  `// Function to return sum ` `static` `int` `sum(``int` `k, ``int` `n) ` `{ ` `    ``int` `sum = (``int``)(Math.pow(k, n + ``1``) -  ` `                    ``Math.pow(k - ``1``, n + ``1``)); ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `n = ``3``; ` `    ``int` `K = ``3``; ` `    ``System.out.print(sum(K, n)); ` `} ` `} ` ` `  `// This code is contributed ` `// by Akanksha Rai `

## Python3

 `# Function to return sum  ` `def` `sum``(k, n): ` `    ``sum` `=` `(``pow``(k, n ``+` `1``) ``-`  `           ``pow``(k ``-` `1``, n ``+` `1``));  ` ` `  `    ``return` `sum``;  ` ` `  `# Driver code  ` `n ``=` `3``;  ` `K ``=` `3``;  ` `print``(``sum``(K, n));  ` ` `  `# This code is contributed by mits `

## C#

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return sum ` `static` `int` `sum(``int` `k, ``int` `n) ` `{ ` `    ``int` `sum = (``int``)(Math.Pow(k, n + 1) -  ` `                    ``Math.Pow(k - 1, n + 1)); ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `n = 3; ` `    ``int` `K = 3; ` `    ``Console.Write(sum(K, n)); ` `} ` `} ` ` `  `// This code is contributed ` `// by Akanksha Rai `

## PHP

 `

Output:

```65
```

Time Complexity: O( n )

3. Third Approach (Efficient): O(log n)

Note: Time complexity can further be reduced to O(log(n)), by calculating power in log(n) complexity.

Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// Recursive C program to compute modular power ` `int` `exponent(``int` `A, ``int` `B) ` `{ ` `    ``// Base cases ` `    ``if` `(A == 0) ` `        ``return` `0; ` `    ``if` `(B == 0) ` `        ``return` `1; ` ` `  `    ``// If B is even ` `    ``long` `y; ` `    ``if` `(B % 2 == 0) { ` `        ``y = exponent(A, B / 2); ` `        ``y = (y * y); ` `    ``} ` ` `  `    ``// If B is odd ` `    ``else` `{ ` `        ``y = A; ` `        ``y = (y * exponent(A, B - 1)); ` `    ``} ` ` `  `    ``return` `y; ` `} ` ` `  `// Function to return sum ` `int` `sum(``int` `k, ``int` `n) ` `{ ` `    ``int` `sum = exponent(k, n + 1) ` `              ``- exponent(k - 1, n + 1); ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``int` `K = 3; ` `    ``cout << sum(K, n); ` `} `

## Java

 `import` `java.lang.Math; ` ` `  `class` `GFG ` `{ ` `     `  `// Recursive C program to compute modular power ` `static` `int` `exponent(``int` `A, ``int` `B) ` `{ ` `    ``// Base cases ` `    ``if` `(A == ``0``) ` `        ``return` `0``; ` `    ``if` `(B == ``0``) ` `        ``return` `1``; ` ` `  `    ``// If B is even ` `    ``int` `y; ` `    ``if` `(B % ``2` `== ``0``)  ` `    ``{ ` `        ``y = exponent(A, B / ``2``); ` `        ``y = (y * y); ` `    ``} ` ` `  `    ``// If B is odd ` `    ``else`  `    ``{ ` `        ``y = A; ` `        ``y = (y * exponent(A, B - ``1``)); ` `    ``} ` ` `  `    ``return` `y; ` `} ` ` `  `// Function to return sum ` `static` `int` `sum(``int` `k, ``int` `n) ` `{ ` `    ``int` `sum = exponent(k, n + ``1``) ` `            ``- exponent(k - ``1``, n + ``1``); ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``3``; ` `    ``int` `K = ``3``; ` `    ``System.out.println(sum(K, n)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech. `

## Python3

 `# Recursive python3 program to compute modular power ` ` `  `def` `exponent(A, B): ` `    ``# Base cases ` `    ``if` `(A ``=``=` `0``): ` `        ``return` `0``; ` `    ``if` `(B ``=``=` `0``): ` `        ``return` `1``; ` ` `  `    ``# If B is even ` `    ``if` `(B ``%` `2` `=``=` `0``): ` `        ``y ``=` `exponent(A, B ``/` `2``); ` `        ``y ``=` `(y ``*` `y); ` ` `  `    ``# If B is odd ` `    ``else``: ` `        ``y ``=` `A; ` `        ``y ``=` `(y ``*` `exponent(A, B ``-` `1``)); ` ` `  `    ``return` `y; ` ` `  `# Function to return sum ` `def` `sum``(k, n): ` `    ``sum` `=` `exponent(k, n ``+` `1``) ``-` `exponent(k ``-` `1``, n ``+` `1``); ` ` `  `    ``return` `sum``; ` ` `  `# Driver code ` `n ``=` `3``; ` `K ``=` `3``; ` `print``(``sum``(K, n)); ` ` `  `# This code has been contributed by 29AjayKumar `

## C#

 `// C# program of above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Recursive C program to compute modular power ` `static` `int` `exponent(``int` `A, ``int` `B) ` `{ ` `    ``// Base cases ` `    ``if` `(A == 0) ` `        ``return` `0; ` `    ``if` `(B == 0) ` `        ``return` `1; ` ` `  `    ``// If B is even ` `    ``int` `y; ` `    ``if` `(B % 2 == 0)  ` `    ``{ ` `        ``y = exponent(A, B / 2); ` `        ``y = (y * y); ` `    ``} ` ` `  `    ``// If B is odd ` `    ``else` `    ``{ ` `        ``y = A; ` `        ``y = (y * exponent(A, B - 1)); ` `    ``} ` ` `  `    ``return` `y; ` `} ` ` `  `// Function to return sum ` `static` `int` `sum(``int` `k, ``int` `n) ` `{ ` `    ``int` `sum = exponent(k, n + 1) ` `            ``- exponent(k - 1, n + 1); ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `n = 3; ` `    ``int` `K = 3; ` `    ``Console.WriteLine(sum(K, n)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech. `

## PHP

 ` `

Output:

```65
```

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