F(x) = 1/√5( ((1+√5)/2)x - ((1-√5)/2)x )
prove that in series S
f(1),f(2)............f(n);
[f(k)] , [f(k+2)], [f(k+3)] are in A.P.
where k and n ε N and k
1not too difficult just try to analyse the function or give it a guess who knows it might be correct can't give any hint bcoz even an air of a hint will ruin the ques
1the point is how do we bring recursive( or recurrence relations ) into their closed form?
just by staring at them for long enough or there is some other way?
this came to mind after seeing 12-01-09 QOTD
P.S. = staring means after staring, testing their valus for arbitrary data and using induction
62Evenidontknow philip.. ..
i do it by staring at it.. and it generally works for me.....
I dont know of a "general" method..
1could you bring the above function to its closed form or did you read of it somewhere like me
ok then how should we approach your QOTD for 12/1?
by staring [7]
62philip this is doable if u take this as
an-(1-a)n
Otherwise it is very tough to guess :(
9hey i wonder y u gave the [] fn cos range of f is always integer
after that its easy to prove
9f is of frm 1/√5 (Yx-(-1/Y)x)
were Y2-Y-1=0
Y - 1/Y =1
Y +1/Y=√5
using abv two and some logic in taking exponents on both sides u can prove that f is always integer so [] is uneccasry
thereon its easy to prove they r in ap by eqating that fx + fx+3 = 2fx+2
1i don't think u got what i wanted u to get celestine
but yes good work
though you already know that it is in AP if you can just look at the question more carefully
and yes sorry about the [] but anyway i don't think that makes any difference
1well celes the point is this is posted in puzzles
the method you are trying will take time
why don' t you find out some terms of this function and see
maybe something popular and familiar would come up [11][1]
9yeah sub for first few terms will give only int but thats not convincing
1ok i don't have any method but this might amuse you
Prove That for fibonacci series :==
the kth (k+2)th and (K+3)th terms are in A.P. [4][11]
9k got it
but its evident to u as its ur q but nt evident frm my point of view