Question

Determine 208/207+208/207×206/205+208/207×206/205×204/203+…

Answer 1

f(208)=208/207+208/207×206/205+208/207×206/205×204/203+…
=208

This can be obtained by induction:
Let f(n) be the function such that:
f(n)=n/(n-1) + n/(n-1)*(n-2)/(n-3) + … n/(n-1)*(n-2)/(n-3)…(2/1)
where n is even.
Then
f(2)=2/1=2
f(4)=4/3+(4/3)*(2/1)=4/3+8/3=4
f(6)=6/5+(6/5)*(4/3)+(6/5)*(4/3)*(2/1)
=6/5(1+(4/3)+(4/3)(2/1))
=6/5(1+(4/3)(1+2/1)
=6/5(1+(4/3)(1+2))
=6/5(1+(4/3)(3))
=(6/5)(1+4)
=6/5(5)
=6

By nesting, and using mathematical induction, we can prove that f(n)=n

Therefore:
f(208)=208/207+208/207×206/205+208/207×206/205×204/203+… = 208