A sequence is defined by u(1) = 1 and u(n+1) = u(n) / [u(n) + 2] for all positive integers of n.
Write down the first five terms of the sequence {1/u(n)}.
my answer: {1, 3, 5/3, 11/5, 21/11}
Hence make a conjecture for u(n) and prove your conjecture using induction.
this is the part where i'm stuck at. how do i go about making the conjecture?
Hi,
Please check your work again.
I got the following:
u_2 = 1/3, so 1/u_2 = 3 = 2^2 - 1;
u_3 = 1/7, so 1/u_3 = 7 = 2^3 - 1;
u_4 = 1/15, so 1/u_4 = 15 = 2^4 - 1;
u_5 = 1/31, so 1/u_5 = 31 = 2^5 - 1;
u_6 = 1/63, so 1/u_6 = 63 = 2^6 - 1.
Hence 1/u_n = 2^? - 1.
The key to making a conjecture is to look at patterns. If you refer to the answers you have obtained, there does not seem to have a pattern, so you may suspect that something is not right and you ought to look back at your working.
Thanks!
Cheers,
Wen Shih
Hi,
The type of conjecture we saw is one of the common types, where it can be expressed as a power of some number, where the power is related to the subscript of u.
Suppose u_1 = 1/2, u_2 = 3/5, u_3 = 2/3, u_4 = 5/7, u_5 = 3/4. This type of conjecture involves a fraction where the numerator and the denominator are related, and they can be expressed in terms of the subscript. If we look closely at the sequence again:
u_1 = 2/4, u_2 = 3/5, u_3 = 4/6, u_4 = 5/7, u_5 = 6/8.
We can now see a pattern more clearly and conclude that u_n = (n + 1) / (n + 3).
I will describe another type in my next posting. Thanks!
Cheers,
Wen Shih
oh so that was where my mistake was...
thanks mr wee!
Hi,
Suppose we have a sequence that looks like this:
u_1 = 5/3, u_2 = 1/3, u_3 = 1/7, u_4 = 5/63, u_5 = 5/99.
We could guess that the numerator is always 5 and the denominator is a product of two factors. The two factors may also be related.
Rewriting the sequence, we obtain
u_1 = 5 / (1)(3),
u_2 = 5 / (3)(5),
u_3 = 5 / (5)(7),
u_4 = 5 / (7)(9),
u_5 = 5 / (9)(11).
It is now possible for us to state the formula
u_n = 5 / (2n - 1)(2n + 1).
Thanks!
P.S. It is useful to remember some results when making your conjecture:
1. 2n is always even.
2. (2n - 1) or (2n + 1) is always odd.
3. (-1)^n = -1 for odd n, while (-1)^n = 1 for even n.
4. n! = n(n - 1)(n - 2)...(3)(2)(1) or n! = n(n - 1)! equivalently.
Cheers,
Wen Shih
Hi,
Suppose we have this sequence:
u_1 = (1)(1^2), u_2 = (2)(4^2), u_3 = (3)(7^2), u_4 = (4)(10^2).
Now look at the second terms without powers, i.e. 1, 4, 7, 10. We can see an arithmetic sequence with first term 1 and common difference 3.
Thus, we are able to make the conjecture u_r = (r) (3r - 2)^2.
Thanks!
P.S. It is helpful to remember AP/GP concepts when making conjectures.
Cheers,
Wen Shih