sum of infinite series..genious or is there a theorem..admission test Thai University

[hebat]

I think it is better to see the attached file because the editor doesn't seem to accept my editing.

a1 a2 a3 infinity an 2^n 3^n
-- + -- + -- + .... = sum --- where an= ------ and bn= -------
b1 b2 b3 n=1 bn n*(n+2) 5 *n+18

(note: a1 is a subscript 1 and an is a subscript n and bn is b subscript 1)

The Thai tutor shows his solution where he eventually comes to a point where he simplifies the sum which make things easy to cancel one and another out.
What gave him the idea to go to that point? How would students know to work it out to that point or is there a theorem I'm missing.

Let me try to show his solution:


an 5*n+18 9*(n+2) - 4*n 9 4
--- = ---------- * (2/3)^n = (----------------)*(2/3)^n = (--- - ------)*(2/3)^n
bn n*(n+2) n*(n+2) n n+2

It took me a couple of minutes to figure out what he was doing. But the step above made me wondering why would he want to do this, but all right...

Then eventually he arrives at this point.

infinity 9 4 infinity 9 infinity 4
sum [ ---*(2/3)^n - --- *(2/3)^n] = sum --- *(2/3)^n - sum ----- * (2/3)^n
n=1 n n+2 n=1 n n=1 n+2

And this is exactly my question how did he know beforehand that this will result to cancel things out. What is the background knowledge to make him go to this point?

= [ 6 + 2 + 8/9 + 4/9 + .... ] - [ 8/9 + 4/9 + ....] = 6 + 2 = 8

finally the answer is EIGHT

Once again my question is if there's a kind of theorem to solve this problem.
Are these questions supposed to be for high school students of grade 12?

Anyway, I would appreciate the support and I thank you in advance.

 

[Ishuda]

I think it is better to see the attached file because the editor doesn't seem to accept my editing.

a1 a2 a3 infinity an 2^n 3^n
-- + -- + -- + .... = sum --- where an= ------ and bn= -------
b1 b2 b3 n=1 bn n*(n+2) 5 *n+18

(note: a1 is a subscript 1 and an is a subscript n and bn is b subscript 1)

The Thai tutor shows his solution where he eventually comes to a point where he simplifies the sum which make things easy to cancel one and another out.
What gave him the idea to go to that point? How would students know to work it out to that point or is there a theorem I'm missing.

Let me try to show his solution:


an 5*n+18 9*(n+2) - 4*n 9 4
--- = ---------- * (2/3)^n = (----------------)*(2/3)^n = (--- - ------)*(2/3)^n
bn n*(n+2) n*(n+2) n n+2

It took me a couple of minutes to figure out what he was doing. But the step above made me wondering why would he want to do this, but all right...

Then eventually he arrives at this point.

infinity 9 4 infinity 9 infinity 4
sum [ ---*(2/3)^n - --- *(2/3)^n] = sum --- *(2/3)^n - sum ----- * (2/3)^n
n=1 n n+2 n=1 n n=1 n+2

And this is exactly my question how did he know beforehand that this will result to cancel things out. What is the background knowledge to make him go to this point?

= [ 6 + 2 + 8/9 + 4/9 + .... ] - [ 8/9 + 4/9 + ....] = 6 + 2 = 8

finally the answer is EIGHT

Once again my question is if there's a kind of theorem to solve this problem.
Are these questions supposed to be for high school students of grade 12?

Anyway, I would appreciate the support and I thank you in advance.

The idea for manipulating the given series is from a lot of work doing problems like this. Looking at your equation (1), one will, with enough experience, recognize this as a difference in two infinite series with (possibly) different constants multipling then. That is let S be the sum of the infinite series we don't know yet but will derive and Sum be the answer we are looking for. Then

. . . . .Sum = c + d S - e S = c + (d-e) S

The rest of the work is manipulations to find the a, b, c, and S: Starting with

. . . . .Sum = \(\displaystyle \displaystyle \sum_{n\,=\,1}^{\infty}\,\dfrac{9}{n}\, \left(\dfrac{2}{3}\right)^n\, -\, \sum_{n\,=\,1}^{\infty}\,\dfrac{4}{n\,+\,2}\, \left(\dfrac{2}{3}\right)^n\)

we note that if we multiply that second summation by

. . . . .\(\displaystyle \left(\dfrac{2}{3}\right)^2\, =\, \dfrac{4}{9}\)

we almost have the same thing as in the first summation. So, lets work on that second summation

. . . . .\(\displaystyle \displaystyle \begin{align} \sum_{n\,=\,1}^{\infty}\, \dfrac{4}{n+2}\, \left(\dfrac{2}{3}\right)^n\, &=\, \dfrac{9}{4}\, \sum_{n\,=\,1}^{\infty}\, \dfrac{4}{n+2}\, \left(\dfrac{2}{3}\right)^{n+2}

\\ \\ &=\, \sum_{n\,=\,1}^{\infty}\, \dfrac{9}{n+2}\, \left(\dfrac{2}{3}\right)^{n+2} \end{align}\)

Letting k=n+2, we have k=3 when n=1 and k=infinity when n=infinity, so

. . . . .= \(\displaystyle \displaystyle \sum_{n\,=\,1}^{\infty}\, \dfrac{4}{n+2}\, \left(\dfrac{2}{3}\right)^n\, =\, \sum_{k\,=\,3}^{\infty}\, \dfrac{9}{k}\, \left(\dfrac{2}{3}\right)^{k}\)

and we have

. . . . .S = \(\displaystyle \displaystyle \sum_{k\,=\,3}^{\infty}\, \dfrac{9}{k}\, \left(\dfrac{2}{3}\right)^{k}\)

We also have d = e = 1, and all we have to do is find c [note, at this point, one should say a word or two about the convergence of S.] Well, c is just what is left over from the first sum above comprising Sum; that is

. . . . .Sum = c = \(\displaystyle \displaystyle \sum_{n\,=\,1}^{n=2}\, \dfrac{9}{n}\, \left(\dfrac{2}{3}\right)^n\, = 9 \left[\dfrac{2}{3}\, +\, \dfrac{1}{2}\, \left(\dfrac{2}{3}\right)^2\right]\, =\, 6\, +\, 2\, =\, 8\)

 

[hebat]

The idea for manipulating the given series is from a lot of work doing problems like this. Looking at your equation (1), one will, with enough experience, recognize this as a difference in two infinite series with (possibly) different constants multipling then. That is let S be the sum of the infinite series we don't know yet but will derive and Sum be the answer we are looking for. Then

. . . . .Sum = c + d S - e S = c + (d-e) S

The rest of the work is manipulations to find the a, b, c, and S: Starting with

. . . . .Sum = \(\displaystyle \displaystyle \sum_{n\,=\,1}^{\infty}\,\dfrac{9}{n}\, \left(\dfrac{2}{3}\right)^n\, -\, \sum_{n\,=\,1}^{\infty}\,\dfrac{4}{n\,+\,2}\, \left(\dfrac{2}{3}\right)^n\)

we note that if we multiply that second summation by

. . . . .\(\displaystyle \left(\dfrac{2}{3}\right)^2\, =\, \dfrac{4}{9}\)

we almost have the same thing as in the first summation. So, lets work on that second summation

. . . . .\(\displaystyle \displaystyle \begin{align} \sum_{n\,=\,1}^{\infty}\, \dfrac{4}{n+2}\, \left(\dfrac{2}{3}\right)^n\, &=\, \dfrac{9}{4}\, \sum_{n\,=\,1}^{\infty}\, \dfrac{4}{n+2}\, \left(\dfrac{2}{3}\right)^{n+2}

\\ \\ &=\, \sum_{n\,=\,1}^{\infty}\, \dfrac{9}{n+2}\, \left(\dfrac{2}{3}\right)^{n+2} \end{align}\)

Letting k=n+2, we have k=3 when n=1 and k=infinity when n=infinity, so

. . . . .= \(\displaystyle \displaystyle \sum_{n\,=\,1}^{\infty}\, \dfrac{4}{n+2}\, \left(\dfrac{2}{3}\right)^n\, =\, \sum_{k\,=\,3}^{\infty}\, \dfrac{9}{k}\, \left(\dfrac{2}{3}\right)^{k}\)

and we have

. . . . .S = \(\displaystyle \displaystyle \sum_{k\,=\,3}^{\infty}\, \dfrac{9}{k}\, \left(\dfrac{2}{3}\right)^{k}\)

We also have d = e = 1, and all we have to do is find c [note, at this point, one should say a word or two about the convergence of S.] Well, c is just what is left over from the first sum above comprising Sum; that is

. . . . .Sum = c = \(\displaystyle \displaystyle \sum_{n\,=\,1}^{n=2}\, \dfrac{9}{n}\, \left(\dfrac{2}{3}\right)^n\, = 9 \left[\dfrac{2}{3}\, +\, \dfrac{1}{2}\, \left(\dfrac{2}{3}\right)^2\right]\, =\, 6\, +\, 2\, =\, 8\)

Thank you so much to explain this.
I guess I can figure out the manipulation part. I haven't yet studied infinite series and how to re-index series that much but if it is a common exercise to recognise the start of the equation as a difference of two infinite series then I will just to work on that.
But still I'm having doubts if these exercises are common for high school students grade 12. I mean to recognise the start equation as a difference of two infinite series.

Once again thank you so much.