Evaluate this sum in terms of n (the number of terms):
1 2 4 8 16
S = --- + ----- + ----- + ------- + ---------- + .....
2 5 41 3,281 21,523,361
(In reply to
re: Solution by pcbouhid)
IF (as previous writers have been supposing) the nth term, a(n) is given by a(n) = 2^n/(1 + 3^2^(n - 1)) we claim for n>=1 that
Sum[2^k/(1 + 3^2^(k - 1)), {k, 1, n}] = 1 - 2^(1 + n)/( 3^2^n - 1) = s(n)
Then for n = 1
Sum[2^k/(1 + 3^2^(k - 1)), {k, 1, 1}] = 1/2 = 1 - 2^(n + 1)/( 3^2^n - 1) = s(1)
Having established a basis, we assume equality for n = k -1: i.e.
Sum[2^k/(1 + 3^2^(k - 1)), {k, 1, k - 1}] = s(k -1)
Then for n =k, it is fairly straightforward to show
Sum[2^k/(1 + 3^2^(k - 1)), {k, 1, k}] = a(k) + s(k-1) = s(k)
So the claim is true by induction.
Edited on December 22, 2005, 11:50 am
|
|
Posted by goFish
on 2005-12-22 11:47:12 |
re(2): Solution
