solve (2000! / 1000!) = k(1 × 3 × 5 × 7 × ... × 1999) for k
- Thread starter 21385
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Re: Another question
Hello, 21385!
A tricky problem . . . I had to baby-talk my way through it.
We have: \(\displaystyle \
1000!)k\;=\;\frac{2000!}{1\cdot3\cdot5\,\cdots\,1999}\)
On the right side, all the odd factors cancel out.
. . and we have: \(\displaystyle \1000!)k \;=\;2\cdot4\cdot6\,\cdots\,2000\)
Then: \(\displaystyle \:k\;=\;\frac{2\cdot4\cdot6\,\cdots\,2000}{1000!} \;=\;\frac{2\cdot4\cdot6\,\cdots\,2000}{1\cdot2\cdot3\,\cdots\,1000}\)
Watch this: \(\displaystyle \;k\;=\;\frac{\not2^2\cdot\not4^2\cdot\not6^2\cdot\not8^2\,\cdots\,\sout{2000}^2}{\not1\,\cdot\,\not2\,\cdot\,\not3\,\cdot\,\not4\,\cdots\,\sout{1000}} \;=\;2^{^{1000}}\)
Hello, 21385!
A tricky problem . . . I had to baby-talk my way through it.
We have: \(\displaystyle \
1000!)k\;=\;\frac{2000!}{1\cdot3\cdot5\,\cdots\,1999}\)On the right side, all the odd factors cancel out.
. . and we have: \(\displaystyle \
1000!)k \;=\;2\cdot4\cdot6\,\cdots\,2000\)Then: \(\displaystyle \:k\;=\;\frac{2\cdot4\cdot6\,\cdots\,2000}{1000!} \;=\;\frac{2\cdot4\cdot6\,\cdots\,2000}{1\cdot2\cdot3\,\cdots\,1000}\)
Watch this: \(\displaystyle \;k\;=\;\frac{\not2^2\cdot\not4^2\cdot\not6^2\cdot\not8^2\,\cdots\,\sout{2000}^2}{\not1\,\cdot\,\not2\,\cdot\,\not3\,\cdot\,\not4\,\cdots\,\sout{1000}} \;=\;2^{^{1000}}\)