Require solution to this problem.
I want to sum the terms in following progressive manner,
Can you give the ultimate equation by which i can calculate the result of this equation for n=1, 2,3,4,5,6, etc.
x+(xy)+(xyy)+(xyyy)+(xyyyy)+........n times
Mathematics
 nicolas.patrois
 Posts: 117
 Joined: Fri Jul 26, 2013 3:54 pm
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Re: Mathematics
Easy for x, and for y, consider 1+2+3+4+…
Re: Mathematics
First of all, it is not an equation. But if you are intrested in the closed form of that series:
x*n  n*(n1)*y/2
x*n  n*(n1)*y/2

 Posts: 1
 Joined: Sun May 11, 2014 7:02 pm
Re: Mathematics
Let us define the m^{th} term of this sequence:
t_{m} = xyyy...(m times) = xy(1+1+1+1+...(m times)) = xmy
S_{n} = Sum(m=1 to n){t_{m}}
= Sum(m=1 to n){xmy}
= Sum(m=1 to n){x}  Sum(m=1 to n){my}
= x*Sum(m=1 to n){1}  y*Sum(m=1 to n){m} as x and y are constant over all values of m
= x*n  [frac]y*n*(n+1),2[/frac]
t_{m} = xyyy...(m times) = xy(1+1+1+1+...(m times)) = xmy
S_{n} = Sum(m=1 to n){t_{m}}
= Sum(m=1 to n){xmy}
= Sum(m=1 to n){x}  Sum(m=1 to n){my}
= x*Sum(m=1 to n){1}  y*Sum(m=1 to n){m} as x and y are constant over all values of m
= x*n  [frac]y*n*(n+1),2[/frac]