Olympiad Maths Prep

Track / Stage 4 / 251 of 340 #511 of 2000

Problem 511

AMC 12 late, AIME early
Algebra Difficulty 4.9 Find the answer

7. The function ff defined on the set of positive integers N\mathbf{N}^{-} satisfies: f(1)=1f(1)=1, and for any natural numbers m,nm, n, f(m)+f(m)+ f(n)=f(m+n)mnf(n)=f(m+n)-m n, then f(m)=f(m)= \qquad .

Official solution

7. m(m+1)2\frac{m(m+1)}{2} Let n=1,f(m+1)f(m)=m+1n=1, f(m+1)-f(m)=m+1, then use partial fraction decomposition.

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