Please help to solve these problems :

1. (1+2/1) x (1+2/2) x (1+2/3) x .... x (1+2/26) x (1+2/27) =

2. (1+1/4) x (1+1/9) x (1+1/16) x ... x (1+1/4028049) =

Thank you(Speechless)

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- April 15th 2008, 05:22 PMagus hendrofraction problems
Please help to solve these problems :

1. (1+2/1) x (1+2/2) x (1+2/3) x .... x (1+2/26) x (1+2/27) =

2. (1+1/4) x (1+1/9) x (1+1/16) x ... x (1+1/4028049) =

Thank you(Speechless) - April 16th 2008, 05:27 AMPeritus
(1+2/1) x (1+2/2) x (1+2/3) x .... x (1+2/26) x (1+2/27) =

2. (1+1/4) x (1+1/9) x (1+1/16) x ... x (1+1/4028049) =

let's denote the following product by:

thus we can come up with the following recurrence relation for the product:

so we can find iteratively.

- April 16th 2008, 03:22 PMSoroban
Hello, agus hendro!

Here's the first one . . .

Quote:

We have . .

- April 16th 2008, 06:19 PMagus hendrofraction problem correction
The last two problem i've posted has been solved the first one. Thanks to Peritus and Soroban. But I should correct the secondsecond like this :

2. (1 - 1/4) (1 - 1/9) (1 - 1/16) .... (1 - 1/2007^2).

I think there is a simple solution like the first problem. Please help. Thank you (Evilgrin) - April 17th 2008, 02:17 AMPeritus
(1 - 1/4) (1 - 1/9) (1 - 1/16) .... (1 - 1/2007^2) =