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f:N-->N
f(0) = 1
, n>=1
i.e.
1 + ((0*f*0) + (1*f*1) + (2 * f *2) ....(n-1)*f*(n-1)) <-- I'm sure this isn't right
then f(4) =
1+ ((0*4*0) + (1*4*1) + (2 * 4 *2) + (3 * 4 *3))
= 1 + 0 + 4 + 16 + 36
= 55
Have I worked this correctly?
Thanks for your time.
No doesn't appear correct
How about this recurrence based on your defintion
f(n) = n.f(n-1)
Am I correct in assuming f stands for function, f is not a numeric variable. only the n inside the function(n-1) is substituted for a number.
f(n) = n.f(n-1) < -- ok, I see, that makes sense. The the final value goes up to n(n-1)
so, how about
1 + ( 1(f(n-1)) + (2(f(n-1)) ....(n(f(n-1)))
then f(4) =
1+ 1(1-1) + 2(2-1) + 3(3-1) + 4(4-1)
= 1+ 0 + 2 + 6 + 12
= 21