Alpha writes the infinite arithmetic sequenceBeta writes the infinite geometric sequenceGamma makes a sequence whose term is the

How about the following:

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

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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

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

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     

Alpha writes the infinite arithmetic sequence

Beta writes the infinite geometric sequence

Gamma makes a sequence whose  term is the product of the  term of Alpha's sequence and the  term of Beta's sequence:

What is the sum of Gamma's entire sequence ?

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

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

Thanks Guys! 

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)]*r^n-1    to     [b₁(a₁ - d)] / (1 - r)    ???

Thanks in advance.....

How did you get from [b₁(a₁ - d)]*r^n-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.

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

Thanks Heureka