If F(1)=2 and F(n)=F(n-1)+1/2 for all integers n>1,then F(101)=?

Answer 1

To go from 1 to 101 requires 100 steps. Each step adds 1/2. So, F(101) = F(1) +100*(1/2)= 2+50 = 52.

Answer 2

It is a classic case of the so-called "telescoping sums"!

F(101)-F(100) = 1/2
F(100) - F(99)=1/2
F(99)-F(98)=1/2
...

F(2)-F(1)=1/2

Add all these equations, and note that the 2nd term in LHS of each equation cancels out in addition with the 1st term in the LHS of the next, so that the sum "telescopes" down to F(101)-F(1) and the right hand side obviously is 1/2 times the number of equations, which is 100.

So F(101)=F(1)+1/2*100=50+2=52

Answer 3

F(1) = 2
F(n) = F(n-1) + 1/2

F(2) = F(2-1) + 1/2 = F(1) + 1/2 = 2 + 1/2 = 2 1/2 = 1/2*2 +1 1/2
F(3) = F(3-1) + 1/2 = F(2) + 1/2 = 2 1/2 + 1/2 = 3 = 1/2*3 + 1 1/2
.
.
.
F(101) = 1/2*101+1 1/2 = 52

Answer 4

52... this is a recursive function, the most famous example of which is the fibonacci series. since F(1)=2, and F(n)=F(n-1)+1/2, F(2)=F(1)+1/2=2+1/2=5/2.
F(3)=5/2+1/2=3... and so on and so forth

Answer 5

umm, 100.5