Sequences series math help

1. Jun 17, 2006

masterofthewave124

For the geometric sequence with tn = 2(-1)^n*(1/3)^n
(a) the sum of the first 99 terms
(b) the sum of the odd-numbered terms t1 + t3 + t5 +...+ t99
(c) the sum of the even-numbered terms t2 + t4 + t6 +...+ t98

so do i first maybe want to convert that into something simpler? why would they write it like that in the first place. a = -2/3, r = -1/3 so wouldnt
tn = (-2/3)(-1/3)^(n-1)?

anyways, after doing that, should i turn that equation into Sn = ?

how would i do b) and c)?

2. Jun 17, 2006

HallsofIvy

Staff Emeritus
Yes, the general term can be written either way: or even as 2(-1/3)^n which I would consider simpler still. Now use the formula to find S99[/b].

For b and c, write out a few terms and see if you can't write those also as a geometric series with a different a and r.

3. Jun 17, 2006

masterofthewave124

ok, i got -1/2, -3/4 and 1/4 for a), b) and c) respectively...are these rounded numbers? simply because as n approaches infinity, the terms become more rational? i see that the relationship is the sum of the sums found in b) and c) equals the sum found in a) so i guess its correct (i..e -3/4 +1/4 =1/2). now what would be the sum of the reciprocals of the terms for the first 99 terms? for this one, a = -3/2 and r = -3 so Sn = (3/8)*((-3)^n-1) correct? but the sum this time is not a nice number, i get -6.44 x 10^46, can anyone verify that?

i have another question;

use mathematical induction to prove the following:

http://img90.imageshack.us/img90/7200/induction2li.jpg [Broken]

can this be restated as this:

n!/(n-0)!0! + n!/(n-1)!1! + n!/(n-2)!2! + n!/(n-r)!r! = 2^n?

how would i prove true for n = k+1?

i have
LS = 2^k + (k+1)!/(k+1-r)!r!
RS = 2^(k+1)

are those correct and if so, where do i go from here?

Last edited by a moderator: May 2, 2017
4. Jun 18, 2006

VietDao29

Uhmmm, not pretty sure if it can be Proven by Induction. However, if you notice that 2n = (1 + 1)n, and use the Bonimial Expansion, the problem works out nicely. :)
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And no, it should not be -1/2, -3/4, and 1/4 for the firstproblem. They are all rounded up, it should be some rounding error with the calculator. -1/2 is the sum of the series as n tends to infinity, not for n = 99.
So:
$$t_n = 2 \times \left( - \frac{1}{3} \right) ^ n$$
$$S_n = \sum_{k = 1} ^ n t_n$$
So the sum of the first 99 terms of the series is:
$$S_{99} = -\frac{2}{3} \times \frac{\left( - \frac{1}{3} \right) ^ {99} - 1}{- \frac{1}{3} - 1} = -\frac{2}{3} \times \frac{ - \frac{1}{3 ^ {99}} - 1}{- \frac{4}{3}} = \frac{1}{2} \times \left( - \frac{1}{3 ^ {99}} - 1 \right)$$
You can just do the same for b, and c.
Can you go from here? :)

5. Jun 18, 2006

Curious3141

Find a non-inductive algebraic proof of [k,r] +[k,(r-1)] = [(k+1),r]. This shouldn't be too difficult, just simplify and bring everything under one denominator.

After you get that result, the rest is easy.

$$2^{(k+1)} = 2(2^k) = 2 \sum_{r=0}^{k}[r,k] = 2S(k)$$

Where S(k) represents the summation. Expand out the summation term-wise and instead of simply multiplying by two, write it in the form S(k) + S(k). Add in a "staggered" fashion - the second term of the left hand sum to the first of the right hand sum and use the previously proven result to simplify. The remaining two terms can be expressed in a similar fashion to fit in with the form of the series. You will find that the result is of the required form S(k+1), proving the result.

This, in fact, is how you would go about proving the Binomial Theorem for natural number exponents using induction. I'm simply adapting the method to this. BTW, I don't think you can assume the Binomial Theorem and just write down the answer as VietDao suggests, I believe the question is asking for a more involved proof like the one I've suggested.

Last edited by a moderator: May 2, 2017