1. ## fraction 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

2. (1+2/1) x (1+2/2) x (1+2/3) x .... x (1+2/26) x (1+2/27) = $\prod\limits_{k = 1}^{27} {\left( {1 + \frac{2}
{k}} \right) = } \prod\limits_{k = 1}^{27} {\left( {\frac{{k + 2}}
{k}} \right) = } \frac{{\frac{{29!}}
{{2!}}}}
{{27!}} = \frac{{28*29}}
{2} = 14*29$

2. (1+1/4) x (1+1/9) x (1+1/16) x ... x (1+1/4028049) = $\prod\limits_{k = 2}^{2007} {\left( {1 + \frac{1}
{{k^2 }}} \right) = } \prod\limits_{k = 2}^{2007} {\left( {\frac{{1 + k^2 }}
{{k^2 }}} \right) = } \frac{{\prod\limits_{k = 2}^{2007} {\left( {1 + k^2 } \right)} }}
{{\left( {2007!} \right)^2 }}$

let's denote the following product by:

$
P_m = \prod\limits_{k = 2}^m {\left( {1 + k^2 } \right)}
$

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

$
\left\{ \begin{gathered}
P_{m + 1} = P_m \left[ {1 + \left( {m + 1} \right)^2 } \right] \hfill \\
P_2 = 5 \hfill \\
\end{gathered} \right.
$

so we can find $P_{2007}$ iteratively.

$
\text{(1 + 1/4) x (1 + 1/9) x (1 + 1/16) x }...\text{ x (1 + 1/4028049) = }\frac{{P_{2007} }}
{{\left( {2007!} \right)^2 }}
$

3. Hello, agus hendro!

Here's the first one . . .

$1)\;\; \;=\;\left(1+\frac{2}{1}\right)\cdot\left(1+\frac{ 2}{2}\right)\cdot\left(1+\frac{2}{3}\right) \hdots \left(1+\frac{2}{26}\right)\cdot\left(1+\frac{2}{2 7}\right)\; =$

We have . $S \;=\;\frac{3}{1}\cdot\frac{4}{2}\cdot\frac{5}{3}\c dot\frac{6}{4}\cdot\frac{7}{5}\cdots\frac{28}{26}\ cdot\frac{29}{27}$ . $= \;\frac{\frac{29!}{2}}{27!}\;=\;\frac{29\cdot28}{2 } \;=\;\boxed{406}$

4. ## fraction 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).
5. (1 - 1/4) (1 - 1/9) (1 - 1/16) .... (1 - 1/2007^2) = $\prod\limits_{k = 2}^{2007} {\left( {1 - \frac{1}