network profile
| bio | website | flavors.me/clschnei |
|---|---|---|
| location | Saginaw, MI | |
| age | 24 | |
| visits | member for | 1 year, 6 months |
| seen | Dec 5 '11 at 20:45 | |
| stats | profile views | 17 |
Computer Science student at SVSU.
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Dec 5 |
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How many strings are there of $4$ or fewer lower case letters that have the letter '$x$' in them? Thank you, I know this is a relatively facile for you guys so I appreciate the explanation. |
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Dec 5 |
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How many strings are there of $4$ or fewer lower case letters that have the letter '$x$' in them? Shouldn't the 2 letter calculation be 26^2 - 25^2 (top solution for repeating x being allow)? |
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Dec 5 |
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How many strings are there of $4$ or fewer lower case letters that have the letter '$x$' in them? Yes repeating x is allowed. |
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Nov 14 |
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Show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer Thank you for the prompt answer, by the way. |
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Nov 14 |
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Show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer They only can help if they would respond to their emails. Otherwise, I wait until tomorrow. Yet another case where math professors at my university are essentially useless. haha |
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Nov 14 |
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Show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer Yes, it is Fibonacci. Doesn't claim on the assinment, but does reference a set of problems from the book. My mistake, I apologize. |
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Nov 14 |
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Show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer It doesn't say...so I have no answer for that. |
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Oct 31 |
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Use congruences to show that $6$ divides $n^3 – n$ for every integer $n$ Thanks guys. This discrete math...nothing but hard times. |
