i need to follow

[sujoy]

Respected Sir,
I fail to understand the following step , please explain it lucidly::

f(x)= (x+1)/x * (x+2)/(x+1) * (x+3)/(x+2).....(x+n+1)/(x+n)

=(x+n+1)/x
[Where "n" is a positive integer]
thanks& regards
Sujoy

 

[tkhunny]

For n = 2

f(x) = [(x+1)/x]*[(x+2)/(x+1)]*[(x+3)/(x+2)]

Rearrange (Collect numerators and denominators)

f(x) = [(x+1)*(x+2)*(x+3)]/[(x)*(x+1)*(x+2)]

Rearrange (Align matching numerators and denominators)

f(x) = [1/x]*[(x+1)/(x+1)]*[(x+2)/(x+2)]*[(x+3)/1]

Simplify

f(x) = [1/x]*[1]*[1]*[(x+3)/1]
f(x) = [1/x]*[(x+3)/1]
f(x) = (x+3)/(x)

All factors matched up (and went away) EXCEPT the first in the denominator and the last in the numerator.

 

[soroban]

Hello, sujoy!

tkhunny is absolutely correct!
. . Did you take a good look at that expression?

f(x) = (x+1)/x * (x+2)/(x+1) * (x+3)/(x+2) ... (x+n+1)/(x+n) = (x+n+1)/x
where n is a positive integer.

. . . . . . . . . . . . . . . . x + 1 . . x + 2 . . .x + 3 . . . . . . . x + n + 1
We have: . f(x) . = . ------- * -------- * -------- * . . . * -------------
. . . . . . . . . . . . . . . . . .x . . . .x + 1 . . .x + 2 . . . . . . . . .x + n

Can you see that nearly everything cancels?
All that remains is the first denominator (x) and the last numerator (x + n + 1).

 

[sujoy]

RESPECTED SIR
I WANT TO THANK BOTH OF YOU SIR FOR YOUR KIND SERVICE
REALLY IT IS MY FAULT SO AS TO OVERLOOK SUCH A N EASY PROBLEM
I HAVE REALLY LIKRD THE WAY YOU HAVE OPENED MY EYES
WITH A LOT OF RESPECT
THANKS A LOT SIR
SUJOY