所以,這問的就是… 1+2+3+...100=?
((1+100)*100)/2=5050
f(n) = n + f(n-1)f(0) = 0f(1) = 1 + f(1-1) = 1 + f(0) = 1 + 0 = 1f(2) = 2 + f(2-1) = 2 + f(1) = 2 + 1 = 3..以此類推f(1) = 1f(2) = 3 = 1+2f(3) = 6 = 1+2+3f(4) = 10 = 1+2+3+4..f(99)= 4950 = 1+2+3+....+98+99 = [(1+99)*99]/2f(100) = 100 + f(100-1) = 100 +f(99) = 100 + 4950 = 5050Orf(100) = 1+2+3+......+98+99+100 = [(1+100)*100]/2 = 5050
44有一個遞迴公式,f(n)=n + f(n-1)且f(0)=0,其中n是正整數..-阿摩線上測驗