Maths Question??

Question

If a1=1 and an+1 – 3 an + 2 = 4n then a100 is equal to ____?
NOTE:(1 , n+1, n , 100 appearing after a are in suBscript)

but doug v cant go on solving till a sub100 it will be too tedious to do isnt there a shortcut method of applying sequences and series formulae

Answer 1

Ive been working on this for a few minutes and the closed form expression is bleh

Subscripts by the way are written as such

x_n or x_{n+1} (round brackets work too)

This follows the standard tex rules

EDIT

Closed form rule is (actually it simplified out nice)

a_(n+1)
= 1 + 4 * Sum _ {i=1} ^ n of 3^(i-1) * 4(n-i+1)

EDIT which upon further simplification yields

a_(n+1) = (1-2n)+2 * Sum_{i=1}^n of 3^i

ps if you dont believe me verify it for the first so many values... in fact n = 0 gives a_1 = 1 since the sum is 0 ;)

EDIT I missed an obvious simplification

a_(n+1) = 3^(n+1)-2(n+1)

Which is the simplest possible form

Thus a_100 = a_{99+1} = 3^100-2*100

EDIT

a note to the poster below. You obviously made an error a in your equation should be 1 which would give you the same as me. The factor of 5 should be obvious that it doesnt make sense.

Also in general 'guessing' a possible solution is a horrible way of doing things and does not garuntee it is the unique solution in any way shape or form.

Answer 2

OK, first, I wrote down several of the terms. Then I made a list of the differences. Then I made a list of the differences of those. I noticed that on the 3rd list, each element was 3 times the element before it. What this tells me is, if there were a function that would give me the original sequence as a function of n, then it's 2nd derivative would reduce to something like 3^n.

So, I guessed the original function to be: a*3^n + b*n^2 + c*n + d. Then I formed 4 equations with 4 unknowns using the first 4 terms of the sequence. I solved and got:
a = 5/3, b = 0, c = -2, d = -2.

I went back to the guess: 5*3^(n-1) -2n -2 and found that this does the job, well, at least as far as the first several members of the sequence.

So, a[n] = 5*3^(n-1) - 2n - 2.

Therefore, you get that a[100] = 5*3^(99) - 2(99) -2.

8.58963E43

Answer 3

If I've read correctly what you tried to write, yoou have a sequene of a values (a(sub1), a(sub2), a(sub2), etc.) that follow the recurance relationship

a(sub(n+1)) - 3*a(subn) +2 = 4*n

starting with a(sub1) = 1

then

a(sub2) - 3*a(sub1) + 2 = 4 and a(sub2) = 5

a(sub3) -3*a(sub2) + 2 = 8 and a(sub3) = 21

a(sub4) -3*a(sub3) + 2 = 12 and a(sub4) = 73

a(sub5) -3*a(sub4) + 2 = 16 and a(sub5) = 233

Keep going until you get to a(sub100).

This is called a 'linear difference equation' and there are ways to solve it without all of the arithmetic, but they get pretty messy and involve sumand product notations you probably haven't as yet been exposed to.

Actually, to do the computation(s) use

a(sub(n+1)) = 4*n - 2 + 3*a(subn)

to get each succesive term in the series. Or write a few lines of code (if you do, be sure to use a quad-precision integer for a because it looks as if it's gonna get real big real fast )


Doug

Answer 4

i think i'm getting some of the n's and numbers mixed up...i'll denote subscripts by using parentheses right next to the a's...is that supposed to be a(n+1) -3*a(n+2) = 4*n, with initial condition a(1) = 1? And obviously you're asking what the 100th term is...Because if that's what you're asking, then I would need a second initial condition (i.e., a(2) = ...) in order to solve it. (Also, your answer would involve complex variables if I interpreted the question correctly).

Answer 5

a100=1+99(400+3)=39898