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anonymous
 5 years ago
How do I solve for m,n when f(m,n)=2?
anonymous
 5 years ago
How do I solve for m,n when f(m,n)=2?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The given function was \[f\left ( m,n \right )=\binom{m+n1}{2}+m\] and f(1,1)=1 How do I find what is m and n when f(m,n)=2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the brackets meant "n choose r"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When you look at this thing, you end up with one equation in two unknowns. To solve it, I made some assumptions on the allowable values m and n could take (basically, m and n could be nonnegative integers). Attached is what I worked out for the function after expanding the binomial coefficient. I organised it so that the result would be quadratic in m, for which I then solved. Then, since I assumed only nonnegative integer solutions could be found, it meant only those n for which 8n+9 would be a perfect square, could be taken. From the assumption, then, the lowest n to test would be 0, leading to m = 2 or 1. The next n that would give a perfect square is 5, but for that choice, you end up with m = 1 or 8, and any greater n would lead to more negative solutions for m (since the negative growth in the linear factor outdoes any positive growth in the square root). So we can conclude n=0 and m=2. When you check this, you get\[f(2,0)=(m+n1;2)+m = (2+01;2)+2=(1;2)+2=0+2=2\]as required. I hope this is what you want. Please let me know how it goes :)
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