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Another of Heureka's masterpeices.   (26/1/15)

$$\left[\ 1+\frac{1}{3}+\frac{2}{3}+\frac{2}{9}+\frac{4}{9}+\frac{4}{27}+\frac{8}{27} +\frac{8}{81} +\frac{16}{81}+\dots \ \right] \\\\ \\= 1+ \frac{1}{3} + \frac{2}{3} + \underbrace{\frac{1}{3}*\frac{2}{3} }_{=\frac{2}{9}}+ \underbrace{\frac{2}{3}*\frac{2}{3} }_{=\frac{4}{9}}+ \underbrace{\frac{1}{3}*\frac{4}{3^2} }_{=\frac{4}{27}}+ \underbrace{\frac{2}{3}*\frac{4}{3^2} }_{=\frac{8}{27}}+ \underbrace{ \frac{1}{3}* \frac{8}{3^3}}_{=\frac{8}{81}}+ \underbrace{\frac{2}{3}*\frac{8}{3^3}}_{=\frac{16}{81}}+\dots \ \\\\\\= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{4}{3^2}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{8}{3^3}}+\dots \ \\\\\\= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^2}{3^2}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^3}{3^3}}+\dots \ \\\\\\= 1+ 1+ \frac{2}{3}}+ \frac{2^2}{3^2}}+ \frac{2^3}{3^3}}+\dots \ \\\\\\s=1+1*( \frac{2}{3} ) ^0+1*(\frac{2}{3})^1+1*(\frac{2}{3})^2+1*(\frac{2}{3})^3+1*(\frac{2}{3})^4+\dots \$$

\left[\ 1+\frac{1}{3}+\frac{2}{3}+\frac{2}{9}+\frac{4}{9}+\frac{4}{27}+\frac{8}{27} +\frac{8}{81} +\frac{16}{81}+\dots \ \right] \\\\\\

= 1+ \frac{1}{3} + \frac{2}{3} + \underbrace{\frac{1}{3}*\frac{2}{3} }_{=\frac{2}{9}}+ \underbrace{\frac{2}{3}*\frac{2}{3} }_{=\frac{4}{9}}+ \underbrace{\frac{1}{3}*\frac{4}{3^2} }_{=\frac{4}{27}}+ \underbrace{\frac{2}{3}*\frac{4}{3^2} }_{=\frac{8}{27}}+ \underbrace{ \frac{1}{3}* \frac{8}{3^3}}_{=\frac{8}{81}}+ \underbrace{\frac{2}{3}*\frac{8}{3^3}}_{=\frac{16}{81}}+\dots \ \\\\\\

= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{4}{3^2}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{8}{3^3}}+\dots \ \\\\\\

= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^2}{3^2}}+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^3}{3^3}}+\dots \ \\\\\\

= 1+ 1+ \frac{2}{3}}+ \frac{2^2}{3^2}}+ \frac{2^3}{3^3}}+\dots \ \\\\\\s=1+1*( \frac{2}{3} ) ^0+1*(\frac{2}{3})^1+1*(\frac{2}{3})^2+1*(\frac{2}{3})^3+1*(\frac{2}{3})^4+\dots \

Very nice Heureka     

The two questions are not the same for the bottom question

5x-5=0  or  x+3=0

so x=1  or    x=-3

how does the discount rate fit into the question?

Use this site - I cannot make it easier than this

That would depend on the diameter of the coin  

This sum will rise in value for 10 years with 10% per year. 1000*1.1^10. but I also want to add 100 dollars per year for 10 years and that amount will also rise in value by 10% per year.

Can someone plz help me with a complete formula for this? Thank you.

Well if the 100 is put in at the beginning of each year for 10 years

The the first 100 will grow to 100*1.1^10

The the first 100 will grow to 100*1.1^9

The the last 100 will grow to 100*1.1^1

The addition will be

$$\\100*1.1^1+100*1.1^2+ .....100*1.1^{10}\\
=100(1.1+1.1^2+ ......+1.1^{10})\\\\
$This is a GP a=100*1.1, r=1.1$\\\\
sum = \frac{a(r^n-1)}{r-1}\\\\
sum = \frac{100*1.1(1.1^{10}-1)}{1.1-1}\\\\
sum = \frac{100*1.1(1.1^{10}-1)}{0.1}\\\\$$

So total after 10 years (all deposits are made at the beginning of the relevant year)

Grand Total = $$1000*1.1^{10}+ \frac{100*1.1(1.1^{10}-1)}{0.1}\\\\$$

Anonymous is correct, but just to flesh it out a little:

Momentum is conserved (assuming no friction etc.) so:

Initial momentum = 0

Final momentum of wagon + child after bricks are thrown = (62 + 150 - 6)*velocity

Final momentum of bricks = -6*2  (assuming velocity is negative to the west and positive to the east)

(62+150-6)*velocity - 6*2 = 0

velocity = 6*2/(62+150-6) m/s

$${\mathtt{velocity}} = {\frac{{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{2}}}{\left({\mathtt{62}}{\mathtt{\,\small\textbf+\,}}{\mathtt{150}}{\mathtt{\,-\,}}{\mathtt{6}}\right)}} \Rightarrow {\mathtt{velocity}} = {\mathtt{0.058\: \!252\: \!427\: \!184\: \!466}}$$ 

or velocity ≈ 0.058 m/s

.

$$\\g(x)=4(x-3)^2+1 find g(-2)\\\\
g(-2)=4(-2-3)^2+1 \\\\
g(-2)=4(-5)^2+1 \\\\
g(-2)=4(25)+1 \\\\
g(-2)=100+1 \\\\
g(-2)=101 \\\\$$

 This could also be solved by putting p(t) = 50 and solving for t.  There will be two values for t (one for the ball on its way up and one for it on its way down).  The difference between them gives the time you are looking for.

The equation can be written as : 50 = -16t² + 50t + 20 or:

16t² - 50t + 30 = 0 

$${\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,\small\textbf+\,}}{\mathtt{30}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{145}}}}{\mathtt{\,-\,}}{\mathtt{25}}\right)}{{\mathtt{16}}}}\\
{\mathtt{t}} = {\frac{\left({\sqrt{{\mathtt{145}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{25}}\right)}{{\mathtt{16}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{0.809\: \!900\: \!338\: \!825\: \!481\: \!5}}\\
{\mathtt{t}} = {\mathtt{2.315\: \!099\: \!661\: \!174\: \!518\: \!5}}\\
\end{array} \right\}$$

Giving the same result as found graphically by Chris

.

$$(3s+4)(3s^{2}-4s-4)$$

Do you want to expand this?

$$\\(3s+4)(3s^{2}-4s-4)\\\\
= 3s(3s^{2}-4s-4)\;\;+4(3s^{2}-4s-4)$$

can you take it from there?

pi is defined as the ratio of the circumference (c) of a circle to its diameter (d).   

pi = c/d

.

$${{\mathtt{2}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}\left({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{18}}\right)\right) = {\mathtt{448}}$$

1000mb . (0.5)8.85/5.6

$${\frac{{\mathtt{1\,000}}{\mathtt{\,\times\,}}\left({\mathtt{0.5}}\right){\mathtt{\,\times\,}}{\mathtt{8.85}}}{{\mathtt{5.6}}}} = {\mathtt{790.178\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

790.18mb

1=(-2/5)(21)+b   simplify

1 = (-42/5)+ b    add (-42/5) to both sides

1 + 42/5 = b

[5 + 42 ] / 5  = b  = 47/5

Use the tangent inverse on the calculator

Hit the "2nd" button, then the "atan" button, then key in your value and then hit the "=" key......

300 million = 3 x 10⁸

So we have

[1 x 10⁹ ] / [ 3 x 10⁸ ] =

(1/3) X 10   =  about 3 numbers per person

If you're referring to trailing zeroes....

The top number has 40 + 8 + 1 = 49 trailing zeroes

The bottom number has 24 + 4  = 28 trailing zeroes

So, the final result will have  [49 - 28]  = 21 trailing zeroes.....!!!!

Here is an explanation of how this "works".....

Congratulations Jillybean    

1/6.2*100

$${\frac{{\mathtt{1}}}{{\mathtt{6.2}}}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\frac{{\mathtt{500}}}{{\mathtt{31}}}} = {\mathtt{16.129\: \!032\: \!258\: \!064\: \!516\: \!1}}$$    percent

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     

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    

Chris drew my attention to this question.  He really liked it and it does look very interesting. 

I have given it a puzzle icon and added it to our puzzle thread.   Thanks Chris    

Thanks Chris,

I am waiting for Heureka or Alan to come along and do it in 3 or 4 lines  LOL   

P.S.

$$\left[\ 1+\frac{1}{3}+\frac{2}{3}+\frac{2}{9}+\frac{4}{9}+\frac{4}{27}+\frac{8}{27} +\frac{8}{81} +\frac{16}{81}+\dots \ \right] \\\\ \\
= 1+ \frac{1}{3} + \frac{2}{3}
+ \underbrace{\frac{1}{3}*\frac{2}{3} }_{=\frac{2}{9}}
+ \underbrace{\frac{2}{3}*\frac{2}{3} }_{=\frac{4}{9}}
+ \underbrace{\frac{1}{3}*\frac{4}{3^2} }_{=\frac{4}{27}}
+ \underbrace{\frac{2}{3}*\frac{4}{3^2} }_{=\frac{8}{27}}
+ \underbrace{ \frac{1}{3}* \frac{8}{3^3}}_{=\frac{8}{81}}
+ \underbrace{\frac{2}{3}*\frac{8}{3^3}}_{=\frac{16}{81}}
+\dots \ \\\\\\
= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{4}{3^2}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{8}{3^3}}
+\dots \ \\\\\\
= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^2}{3^2}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^3}{3^3}}
+\dots \ \\\\\\
= 1+ 1
+ \frac{2}{3}}
+ \frac{2^2}{3^2}}
+ \frac{2^3}{3^3}}
+\dots \ \\\\\\
s=
1+1*( \frac{2}{3} ) ^0
+1*(\frac{2}{3})^1
+1*(\frac{2}{3})^2
+1*(\frac{2}{3})^3
+1*(\frac{2}{3})^4+
\dots \$$