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

• Difficulty Level : Easy
• Last Updated : 05 Apr, 2021

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 using namespace std; int sum(int M, int P){    return 2*(M + P);} // driver codeint main() {     int M = 2, P = 9;       cout << sum(M,P);       return 0;}

## Java

 // Java Program to finding the// sum of the seriesimport 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)) ;    }}

## Python3

 # Python3 Program to finding# the sum of the  series # function that calculate# the sum of the  seriesdef sum_series(M, P):     return int(2 * (M + P)) # Driver functionM = 2P = 9print(sum_series(M ,P))

## C#

 // C# program to finding the// sum of the seriesusing 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)) ;    }}

## PHP

 

## Javascript

 

Output:

22

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