High School Can You Solve This Complex Mathematical Expression Without a Calculator?

[anemone]

Evaluate, without the help of a calculator the following math expression:

$\dfrac{(2\cdot 5+2)(4\cdot 7+2)(6\cdot 9+2)(8\cdot 11+2)\cdots(1998\cdot 2001+2)}{(1\cdot 4+2)(3\cdot 6+2)(5\cdot 8+2)(7\cdot 10+2)\cdots(1997\cdot 2000+2)}$

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[anemone]

Congratulations to the following members for their correct solutions::)

1. laura123
2. soroban
3. kaliprasad
4. eddybob123

I'll show three solutions here because each of them adopted different approach to solve for the problem.

Here is laura123's solution:

Spoiler

$\displaystyle \begin{align*}\dfrac{(2\cdot 5+2)(4\cdot 7+2)(6\cdot 9+2)(8\cdot 11+2)\cdots(1998\cdot 2001+2)}{(1\cdot 4+2)(3\cdot 6+2)(5\cdot 8+2)(7\cdot 10+2)\cdots(1997\cdot 2000+2)}&=\prod_{k=1}^{999}\dfrac{2k(2k+3)+2}{(2k-1)(2k+2)+2}\\&=\displaystyle\prod_{k=1}^{999}\dfrac{4k^2+6k+2}{4k^2+4k-2k-2+2}\\&=\prod_{k=1}^{999}\dfrac{2(k+1)(2k+1)}{2k(2k+1)}\\&=\prod_{k=1}^{999}\dfrac{k+1}{k}\\&=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dots\cdot\dfrac{999}{998}\cdot\dfrac{1000}{999}\\&
=\dfrac{\cancel{2}}{1}\cdot\dfrac{\cancel{3}}{\cancel{2}}\cdot\dfrac{\cancel{4}}{\cancel{3}}\cdot\dots\cdot\dfrac{\cancel{999}}{\cancel{998}}\cdot\dfrac{1000}{\cancel{999}}\\&=1000 \end{align*}$.

Here is soroban's solution:

Spoiler

We have: $\;\dfrac{(12)(30)(56)(90) \cdots (3,\!1998,\!000)}{(6)(20)(42)(72) \cdots (3,\!994,\!002)} $

$\qquad =\;\dfrac{(3\cdot4)(5\cdot6)(7\cdot8)(9\cdot10) \cdots (1999\cdot 2000)}{(2\cdot3)(4\cdot5)(6\cdot7)(8\cdot9) \cdots (1998\cdot1999)}$

$\qquad =\;\dfrac{\dfrac{2000!}{2!}}{\dfrac{1999!}{1!}} \;=\;\dfrac{2000}{2} \;=\;1000$

Here's kaliprasad's solution:

Spoiler

We have $x(x+3) + 2 = (x+1)(x+2)$

So numerator = $3 * 4 * 5\cdots * 2000$

Denominator = $2 * 3 *4\cdots * 1999$

So ratio = $\dfrac{3 * 4 * 5\cdots * 2000}{2 * 3 *4\cdots * 1999}$ = 1000