We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
2776
8
avatar

Alpha writes the infinite arithmetic sequence10, 8, 6, 4, 2, 0 ,\ldots.Beta writes the infinite geometric sequence9, 6, 4, \frac{8}{3}, \frac{16}{9}, \ldots.Gamma makes a sequence whose n^{\text{th}} term is the product of the n^{\text{th}} term of Alpha's sequence and the n^{\text{th}} term of Beta's sequence:10\cdot 9 \quad,\quad 8\cdot 6\quad ,\quad 6\cdot 4\quad,\quad 4\cdot \frac83\quad,\quad 2\cdot \frac{16}{9}\quad,\quad \ldo...What is the sum of Gamma's entire sequence?

 Jan 25, 2015

Best Answer 

 #4
avatar+28123 
+10

How about the following:

 

summations

 

Perhaps I should clarify a step:  Note that n*rn = r*n*rn-1= r*d(rn)/dr  (or perhaps this just muddies the waters - in which case ignore it!)

.

.

 Jan 26, 2015
 #1
avatar+102753 
+5

10,8,6,       AP    a=10   d=-2      Tn=a+(n-1)d

 

$$T_{n+1}=a+nd=10-2n$$   (I have changed n to start at 0)

 

9,6,4,......    GP   a=9     r=2/3       Tn=ar^(n-1)

 

$$\\T_{n+1} = ar^n\\\\
T_{n+1}=9*(\frac{2}{3})^n$$

 

Sum of Gamma's Sequence

 

$$\\S_\infty =\displaystyle\sum_0^\infty\quad (10-2n)*9*\frac{2^n}{3^n}\\\\
S_\infty =\displaystyle\sum_0^\infty\quad 90*\frac{2^n}{3^n}\quad-\quad\displaystyle\sum_0^\infty\quad (18n)*\frac{2^n}{3^n}\\\\
S_\infty =90*\displaystyle\sum_0^\infty\quad \frac{2^n}{3^n}\quad-\quad18\displaystyle\sum_0^\infty\quad \frac{n*2^n}{3^n}\\\\$$

 

     $$\\NOW\\\\
\displaystyle\sum_0^\infty\;\;\frac{2^n}{3^n}
=\displaystyle\sum_0^\infty\;\;\left(\frac{2}{3}\right)^n\\\\
\qquad $ This is a GP a=1 r=2/3$\\\\
=\frac{a}{1-r}=\frac{1}{1-\frac{2}{3}}=\frac{1}{\frac{1}{3}}=3$$

 

$$\\S_\infty =90*\displaystyle\sum_0^\infty\quad \frac{2^n}{3^n}\quad-\quad18\displaystyle\sum_0^\infty\quad \frac{n*2^n}{3^n}\\\\
S_\infty =90*3\;\;-\quad18\displaystyle\sum_0^\infty\quad \frac{n*2^n}{3^n}\\\\
S_\infty =270\;\;-\quad18\displaystyle\sum_0^\infty\quad \frac{n*2^n}{3^n}\\\\$$

 

Here I cheated a little and used Wolfram|Alpha to solve the second sum.  The answer is 6

The hyperlink is below    

Wolfram|Alpha Second sum solution

 

$$\\S_\infty =270\;\;-\quad18\displaystyle\sum_0^\infty\quad \frac{n*2^n}{3^n}\\\\
S_\infty =270-18*6\\\\
S_\infty = 270-108\\\\
S_\infty = 162\\\\$$

.
 Jan 26, 2015
 #2
avatar+102753 
0

If another mathematician can explain to me how to do the second sum solution that would be good.

 

Ie      Why does

 $$\displaystyle\sum_{n\rightarrow0}^\infty\quad \frac{n*2^n}{3^n}=6\qquad ?$$

 

Thank you     

 Jan 26, 2015
 #3
avatar+22855 
+10

Alpha writes the infinite arithmetic sequence10, 8, 6, 4, 2, 0 ,\ldots.

Beta writes the infinite geometric sequence9, 6, 4, \frac{8}{3}, \frac{16}{9}, \ldots.

Gamma makes a sequence whose n^{\text{th}} term is the product of the n^{\text{th}} term of Alpha's sequence and the n^{\text{th}} term of Beta's sequence:10\cdot 9 \quad,\quad 8\cdot 6\quad ,\quad 6\cdot 4\quad,\quad 4\cdot \frac83\quad,\quad 2\cdot \frac{16}{9}\quad,\quad \ldo...

What is the sum of Gamma's entire sequence ?

 

$$\\\small{\text{
Sequence alpha:
$ a_n = a_1 + (n-1)*d = (a_1-d) + n*d
\quad a_1=10$ and $d=8-10=a_{n+1}-a_n=-2$
}}\\
\small{\text{
Sequence beta:
$ b_n = b_1 *r^{n-1} \quad b_1=9$ and $r=\frac{6}{9}=\frac{b_{n+1}}{b_n}=\frac{2}{3}$
}}$\\\\$
\small{\text{
Sequence gamma :
$ g_n = a_n*b_n = b_1 *r^{n-1} [(a_1-d) + n*d]
$
}}\\
\small{\text{
$ \boxed{ g_n = \underbrace{ \left[ b_1(a_1-d) \right] *r^{n-1} }_{sum\ = \dfrac{b_1(a_1-d)}{1-r} } \ +\ b_1d* n r^{n-1} }
$ The sequence of Gamma has two parts.
}}$\\\\$
\small{\text{
The sum of the first part $ \left[ b_1(a_1-d) \right] *r^{n-1} $ is the sum of a geometric sequence $
= \frac{ b_1(a_1-d)}{1-r}
$
}}$\\\\$
\small{\text{
The sum s of the second part $ b_1d* n r^{n-1} $ is:
}} \\
\begin{array}{rcrrrrr}
s & = & (b_1d) * 1 * r^0 +&
(b_1d)* 2 * r^1 \ + &(b_1d) * 3 * r^2 \ +&(b_1d) * 4 * r^3 \ + &(b_1d) * 5 * r^4 \ + \dots \\
r*s & = & & (b_1d)* 1 * r^1 \ + &(b_1d) * 2 * r^2 \ +&(b_1d) * 3 * r^3 \ + &(b_1d) * 4 * r^4 \ + \dots \\
\hline
s-r*s & = & (b_1*d) \ + & (b_1d)*r^1 \ + &(b_1d)*r^2 \ +&(b_1d)*r^3 \ +&(b_1d)*r^4 \ + \dots \\
\end{array}\\
\small{\text{
$
s-r*s = \underbrace{ (b_1*d) \ + (b_1d)*r^1 \ + (b_1d)*r^2 \ +(b_1d)*r^3 \ +(b_1d)*r^4 \ + \dots }_{\text{sum of a geometric sequence }\ = \frac{b_1d}{1-r} }
$
} $\\$
\small{\text{
$
s-r*s = \frac{b_1d}{1-r}
$
}}$\\$
\small{\text{
$
s(1-r) = \frac{b_1d}{1-r}
$
}}$\\\\$
\small{\text{
$
s = \dfrac{b_1d}{(1-r)^2}
$
}}$\\\\$
\small{\text{
The sum of Gamma's sequence
$ = \dfrac{b_1(a_1-d)}{1-r} \ + \dfrac{b_1d}{(1-r)^2}
$
}}$\\\\$
\small{\text{
$ = \left( \dfrac{b_1}{1-r} \right) * \left[ (a_1-d)+\dfrac{d}{(1-r)} \right]
$
}}$$

$$\small{\text{
The sum of Gamma's sequence
$ = \left(
\dfrac{ 9 }{ 1 - \frac{2}{3} }
\right) *
\left[ (10-(-2))+ \dfrac{ (-2) } { 1 - \frac{2}{3} }
\right] $
}}\\
\small{\text{
$ = 9*3 *(12-2*3) = 27*6 = 162
$
}}$$

.
 Jan 26, 2015
 #4
avatar+28123 
+10
Best Answer

How about the following:

 

summations

 

Perhaps I should clarify a step:  Note that n*rn = r*n*rn-1= r*d(rn)/dr  (or perhaps this just muddies the waters - in which case ignore it!)

.

.

Alan Jan 26, 2015
 #5
avatar+102753 
0

Thanks Guys! 

 Jan 27, 2015
 #6
avatar+102320 
0

Heureka....I'm trying to learn more about how to manipulate these series questions....I followed everything except for this.....

How did you get from [b1(a1 - d)]*rn-1    to     [b1(a1 - d)] / (1 - r)    ???

Thanks in advance.....

 

 Jan 27, 2015
 #7
avatar+22855 
+5

How did you get from [b1(a1 - d)]*rn-1    to     [b1(a1 - d)] / (1 - r)    ???

Hi CPhill,

first i tell you about der geometric sequence of beta:

$$\small{\text{
The geometric sequence is
$
b_1 = \underbrace{9*(\frac{2}{3})^0}_{=9}, \ b_2 = \underbrace{9*(\frac{2}{3})^1}_{=6}, \ b_3 = \underbrace{9*(\frac{2}{3})^2}_{=4}, \ b_4 = \underbrace{9*(\frac{2}{3})^3}_{=\frac{8}{3}}, \ \dots \ ,\ b_n = 9*(\frac{2}{3})^{n-1}
$
}} \\
\small{\text{
The sum $s=\frac{9}{1-\frac{2}{3}}$\ , now we set $\ b_1 = 9 $ and $\ r = \frac{2}{3}
$ and see why
}}\\\\
\small{\text{
b_1 = b_1, \ b_2 = b_1 r, \ b_3 = b_1r^2, \ b_4 = b_1r^3, \ \dots \ ,\ b_n = b_1r^{n-1}
$
}} \\
\small{\text{
The sum is
$
s = b_1 + b_1 r + b_1r^2 + b_1*r^3 + b_1r^4 + \ \dots
$
}} \\
\small{\text{
$
\begin{array}{rcll}
\hline
s &=& b_1 \ + & b_1 r + b_1r^2 + b_1*r^3 + b_1r^4 + \ \dots \\
rs &=& & b_1 r + b_1r^2 + b_1*r^3 + b_1r^4 + \ \dots \\
\hline
s-rs &=& b_1 & + 0 + 0 + 0 + 0 +\dots \\
s(1-r) &=& b_1 & \\
s &=& \dfrac{b_1}{1-r} &
\end{array}
$
}} \\
\boxed{
\small{\text{
So, if we have a Geometric Sequence $ b_n = b_1r^{n-1} $ the infinite sum $ s = \dfrac{b_1}{1-r} $
}}
}$$

 

Now we look to Gamma Sequence.  There is a constant:    $$b_1(a_1-d)=g_1$$

  $$\boxed{
\small{\text{
The Geometric Sequence Gamma first part is $ g_{n_{First Part}} = g_1r^{n-1} $ the infinite sum $ s = \dfrac{g_1}{1-r} $
}}
}\\\\
\small{\text{
$ s = \dfrac{g_1}{1-r} =\dfrac{b_1(a_1-d)}{1-r}$
}}\\\\
\small{\text{
set $ b_1 = 9, \ a_1 = 10,\ d = -2, \ r =\frac{2}{3}
$
}}\\
\small{\text{
we have the sum of Gamma Sequence first part:
}}\\
\small{\text{
$
108*(\frac{2}{3})^0 + 108*(\frac{2}{3})^1 + 108*(\frac{2}{3})^2 +108*(\frac{2}{3})^3 + \dots
$
}} \\
\small{\text{
and the infinite sum $s = \frac{108}{1-\frac{2}{3}}$
and $g_{1_{first part}}=108,\ g_{2_{first part}}=108*\frac{2}{3}, \ \dots \ ,\ g_{n_{first part}}=108*(\frac{2}{3})^{n-1} $
}}\\$$

P.S.

$$\\\small{\text{
The finite sum:
$
s_n=a_1 + a_1\cdot r + a_1\cdot r^2 + a_1\cdot r^3 + \dots +a_1\cdot r^{n-1}
$
is
$
s_n = \dfrac{a_1(1-r^n)}{1-r} $
}}\\\\
\small{\text{
The infinite sum:
$
s_n=a_1 + a_1\cdot r + a_1\cdot r^2 + a_1\cdot r^3 +a_1\cdot r^4 \dots
$
}}\\\\
\small{\text{
with
$
s_n = \dfrac{a_1(1-r^n)}{1-r} $
is $s=\lim\limits_{n\to \infty} s_n = \dfrac{a_1}{1-r}$,}}\\\\
\small{\text{
if $\quad -1 }}$$

.
 Jan 27, 2015
 #8
avatar+102753 
0

Thanks Heureka  

 Jan 28, 2015

4 Online Users

avatar