设f(1)=1,f(m+n)=f(m)+f(n)+m*n(n,m都为自然数),f(2007)=?

令 m=1,有
f(n+1)
= f(n)+f(1)+n
= f(n) + (n+1)
故
f(n+1)
= f(n) + (n+1)
= f(n-1) + n + (n+1)
= f(n-2) + (n-1) + n + (n+1)
= ...
= f(1) + 2 + 3 + 4 + ...+ (n+1)
= 1 + 2 + 3 + 4 + ...+ (n+1)
= (n+2)(n+1)/2
f(2007)
= 2008*2007/2
= 1004*2007
= 2015028