# Sum of series M/1 + (M+P)/2 + (M+2*P)/4 + (M+3*P)/8……up to infinite

Find the sum of series M/1 + (M+P)/2 + (M+2*P)/4 + (M+3*P)/8……up to infinite where M and P are positive integers.

Examples:

Input : M = 0, P = 3; Output : 6 Input : M = 2, P = 9; Output : 22

** Method : **

**S = M/1 + (M + P)/2 + (M + 2*P)/4 + (M + 3*P) / 8……up to infinite**

so the solution of this series will be like this

we are going to divide this series into two parts-

S = (M/1 + M/2 + M/4 + M/8……up to infinite) + ( p/2 + (2*p)/4 + (3*p)/8 + ….up to infinite)

let us consider it

**S = A + B ……..eq(1)**

where,

A = M/1 + M/2 + M/4 + M/8……up to infinite

A = M*(1 + 1/2 + 1/4 + 1/8….up to infinite)

which is G.P of infinite terms with r = 1/2;

According to the formula of G.P sum of infinite terms for r < 1 and

a is first term and r is common ratio so now,

A = M * ( 1 / (1 – 1/2) )

A = 2 * M ;

**Now for B –**

B = ( p/2 + (2*p)/4 + (3*p)/8 + ….up to infinite)

B = P/2 * ( 1 + 2*(1/2) + 3*(1/4) + ……up to infinite)

it is sum of AGP of infinite terms with a = 1, r = 1/2 and d = 1;

According to the formula where a is first term,

r is common ratio and d is common difference so now,

B = P/2 * ( 1 / (1-1/2) + (1*1/2) / (1-1/2)^2 )

B = P/2 * 4

B = 2*P ;

put value of A and B in eq(1)

**S = 2(M + P)**

## C++

`#include <iostream> ` `using` `namespace` `std; ` ` ` `int` `sum(` `int` `M, ` `int` `P) ` `{ ` ` ` `return` `2*(M + P); ` `} ` ` ` `// driver code ` `int` `main() { ` ` ` ` ` `int` `M = 2, P = 9; ` ` ` `cout << sum(M,P); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// javaProgram to finding the ` `// sum of the series ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ` `// function that calculate ` ` ` `// the sum of the nth series ` ` ` `static` `int` `sum_series(` `int` `M, ` `int` `P) ` ` ` `{ ` ` ` `return` `2` `* (M + P); ` ` ` `} ` ` ` ` ` `// Driver function ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{ ` ` ` `int` `M = ` `2` `; ` ` ` `int` `P = ` `9` `; ` ` ` `System.out.println( sum_series(M, P)) ; ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

## Python

`# Python 3 Program to finding ` `# the sum of the series ` ` ` `# function that calculate ` `# the sum of the series ` `def` `sum_series(M, P): ` ` ` ` ` `return` `int` `(` `2` `*` `(M ` `+` `P)) ` ` ` `# Driver function ` `M ` `=` `2` `P ` `=` `9` `print` `(sum_series(M ,P)) ` |

*chevron_right*

*filter_none*

## C#

`// C# program to finding the ` `// sum of the series ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `// Function that calculate ` ` ` `// the sum of the nth series ` ` ` `static` `int` `sum_series(` `int` `M, ` `int` `P) ` ` ` `{ ` ` ` `return` `2*(M + P); ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main () ` ` ` `{ ` ` ` `int` `M =2; ` ` ` `int` `P =9; ` ` ` ` ` `Console.Write( sum_series(M,P)) ; ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP program to finding the ` `// sum of the series ` ` ` `// Function that calculate ` `// the sum of the nth series ` `function` `sum(` `$M` `, ` `$P` `) ` `{ ` ` ` `return` `2*(` `$M` `+ ` `$P` `); ` `} ` ` ` `// Driver Code ` `$M` `= 2; ` `$P` `= 9; ` `echo` `sum(` `$M` `, ` `$P` `); ` ` ` `// This code is contributed by mits ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

22

## Recommended Posts:

- Find the sum of infinite series 1^2.x^0 + 2^2.x^1 + 3^2.x^2 + 4^2.x^3 +.......
- Find if the given number is present in the infinite sequence or not
- Minimum moves to reach target on a infinite line | Set 2
- Measure one litre using two vessels and infinite water supply
- Find minimum moves to reach target on an infinite line
- Minimum steps needed to cover a sequence of points on an infinite grid
- Sum of the series Kn + ( K(n-1) * (K-1)1 ) + ( K(n-2) * (K-1)2 ) + ....... (K-1)n
- Sum of the series 1.2.3 + 2.3.4 + ... + n(n+1)(n+2)
- Sum of series 1*1! + 2*2! + ……..+ n*n!
- Sum of the series 1 + (1+2) + (1+2+3) + (1+2+3+4) + ...... + (1+2+3+4+...+n)
- Sum of series 1*1*2! + 2*2*3! + ……..+ n*n*(n+1)!
- Sum of series 1^2 + 3^2 + 5^2 + . . . + (2*n - 1)^2
- Value of the series (1^3 + 2^3 + 3^3 + ... + n^3) mod 4 for a given n
- Sum of the series 2 + (2+4) + (2+4+6) + (2+4+6+8) + …… + (2+4+6+8+….+2n)
- Sum of series (n/1) + (n/2) + (n/3) + (n/4) +.......+ (n/n)

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.