Question

how this interesting pattern works ?
1 * 9 + 2 = 11
12 * 9 + 3 = 111
123 * 9 + 4 = 1111
1234 * 9 + 5= 11111
12345 * 9 + 6 = 111111
123456 * 9 + 7 = 1111111
1234567 * 9 + 8 = 11111111
12345678 * 9 + 9 = 111111111

http://jwilson.coe.uga.edu/EMAT6680/LeeSJ/emat6690/essay1/main.html

Answer 1

do we continue the pattern

Answer 2

cool

Answer 3

thats a good question @jagr2713 , im not sure if we can continue as all the digits are already usedup on left hand side hmm

Answer 4

thats cool

Answer 5

\((0\times 9+1=1)\) \(\leftarrow \text{upper term missing}\)

Answer 6

the last digit of 1, 12, 123, etc all multiplied by 9 become less and less as 2, 3, 4 etc get larger.

Answer 7

the first number (1,12,123,1234 etc) multiplied by 9. add the third number and you get the sequence of 1^x

Answer 8

\(\Large \begin{align*}
\sum _{k=0}^n 10^{n-k} &= \sum _{k=0}^n 10^{n-k} +\sum _{k=0}^n k 10^{n-k}- \sum _{k=0}^n k 10^{n-k} \\
&= \sum _{k=0}^n (k+1)10^{n-k}-\sum _{k=0}^n k 10^{n-k} \\
&= \sum _{k=0}^{n-1} (k+1)10^{n-k}-\sum _{k=0}^n k 10^{n-k}+(n+1)\\
&= 10 \sum _{k=0}^{n} (k)10^{n-k}-\sum _{k=0}^n k 10^{n-k}+(n+1)\\
&= (10-1)\sum _{k=0}^n k 10^{n-k}+(n+1)
\end{align*}\)

Answer 9

What happens when you use binary?

Answer 10

\[
1_b*1001_b+10_b = 1011_b
\]Amazing.

Answer 11

\[
1100_b*1001_b+11_b = 101111_b
\]Unbelievable.

Answer 12

ok i should add this to my solution
number of 1's digits is \(\Large (n+1)\)

Answer 13

thats interesting xD i think it wil take few minutes for me to comprehend the proof

Answer 14

it seems you're starting wid
\[111\ldots (\text{n+1 times}) = \begin{align}
\sum _{k=0}^n 10^{n-k} \end{align}\]

Answer 15

yep

Answer 16

12*9+3 =
11*9+1*8+2+1=
11*9+1+1*9+2
100+11


123*9+4 =
111*9+1+12*9+3
1000+11

I bet induction would do it

Answer 17

WOW @Marki !!!!!! How did you think to discover this?!

Answer 18

:)

Answer 19

This reminds me a lot of how 999 = 9*10^2+9*10^1+9*10^0, I wonder if we can find some larger theory of all of this where both of these results are just easy and obvious to see?

Answer 20

?

Answer 21

what does this means ?
"larger theory of all of this where both of these results are just easy and obvious to see?"

Answer 22

I'm strugglign a bit wid this step :
http://gyazo.com/e404b91fe44bba07363a3cb0d93d6c1a

how did you change the bounds just like that ?

Answer 23

can u see it now ?
\(\begin{align*}
\sum _{k=0}^{n-1}(k+1)10^{n-k} &= 10^n+2.10^{n-1}+3.10^{n-2}+\dots+(n-1)10^2+n10\\
&= 10\left ( 10^{n-1}+2.10^{n-2}+3.10^{n-3}+\dots+(n-1)10+n\right )\\
&= 10 \sum _{k=0}^n k^{n-k}\\
\end{align*}\)

Answer 24

another approach
\(\begin{align*}
\sum _{k=0}^{n-1}(k+1)10^{n-k} &=10\sum _{k=0}^{n-1}(k+1)10^{n-k-1} \\
&= 10\sum _{k=0}^{{\color{Red} {n-1}}}(k+1)10^{{\color{Red} {(n-1)}}-k} ~~~~~\\
&= 10 \sum _{k=0}^{\color{Red} n} k^{{\color{Red} n}-k} \color{Red}{\text{we can sub one variable instead of n-1 }} \\
\end{align*}\)

Answer 25

@ganeshie8

Answer 26

\[\begin{align*}
\sum _{k=0}^{n-1}(k+1)10^{n-k} &=10\sum _{k=0}^{n-1}(k+1)10^{n-k-1} \\
&= 10\sum _{k=0}^{{\color{Red} {n-1}}}(k+1)10^{{\color{Red} {(n-1)}}-k} ~~~~~\\
&= 10 \sum _{k=0}^{\color{Red} n} 10^{{\color{Red} n}-k} \color{Red}{\text{we can sub one variable instead of n-1 }} \\
\end{align*}\]

Answer 27

i made a typo

Answer 28

Makes sense thanks ! im not sure why ganeshie is overthinking :O

Answer 29

\(\begin{align*}
\sum _{k=0}^{n-1}(k+1)10^{n-k} &= 10^n+2.10^{n-1}+3.10^{n-2}+\dots+(n-1)10^2+n10\\
&= 10\left ( 10^{n-1}+2.10^{n-2}+3.10^{n-3}+\dots+(n-1)10+n\right )\\
&= 10 \sum _{k=0}^n k10^{n-k}\\
\end{align*} \)

\(\begin{align*}
\sum _{k=0}^{n-1}(k+1)10^{n-k} &=10\sum _{k=0}^{n-1}(k+1)10^{n-k-1} \\
&= 10\sum _{k=0}^{{\color{Red} {n-1}}}(k+1)10^{{\color{Red} {(n-1)}}-k} ~~~~~\\
&= 10 \sum _{k=0}^{\color{Red} n}k10^{{\color{Red} n}-k} \color{Red}{\text{we can sub one variable instead of n-1 }} \\
\end{align*}\)

like this

Answer 30

hehe ikr @rational ;)

Answer 31

nice nice

Answer 32

:)

Answer 33

just wondering if we can use derivative of sum cleverly here

Answer 34

\[\sum x^k = \dfrac{x^{k+1}-1}{x-1} \implies \sum kx^{k-1} = (\dfrac{x^{k+1}-1}{x-1})'\]

but the k's seem to run in reverse direction in your formula so idk if this works

Answer 35

its work :O

Answer 36

That's what I was working on earlier actually, here's what I got: \[\Large \sum_a^b x^n = \sum_a^bx^{b+a-n}=\frac{x^b-x^a}{x-1}\]

Answer 37

\[\Large \sum_a^b(b+a-n)x^{b+a-n}=\frac{bx^b-ax^a}{x-a}+\frac{x^{a+1}-x^{b-1}}{(x-1)^2}\]

Answer 38

So the first formula has three parts, the first is adding from a to b, the second sum is adding from b to a just like @Marki did, and the right part is the sum of the geometric series, since that's all it is.

The next equation is the derivative, but then I multiplied both sides of the equation by x to get rid of the x^{-1} term.

Then you can also see that the last one I couldn't fit it all on one line: \[\Large \sum_a^b nx^n = \sum_a^b (b+a-n)x^{b+a-n}\] since remember, there were three parts to the first one.

Anyways, have fun playing with this like I did haha. =P

Answer 39

\[\Large \sum_a^b x^n = \sum_a^bx^{b+a-n}=\frac{x^{b+1}-x^a}{x-1} \]

Answer 40

first part is clear, im still going thru second part xD

Answer 41

haha cute

Answer 42

I'm working on rewriting Marki's awesome argument in terms of this more general expression. I'll explain in a second...

Answer 43

Ok here it comes!

Answer 44

\[\large \sum_{k=a}^b x^{a+b-k} = \sum_{k=a}^b x^{a+b-k} + \sum_{k=a}^b kx^{a+b-k}- \sum_{k=a}^b kx^{a+b-k}\] \[\large \sum_{k=a}^b(k+1) kx^{a+b-k}- \sum_{k=a}^b kx^{a+b-k}\]\[\large \sum_{k=a-1}^{b-1}(k+1) kx^{a+b-k}- \sum_{k=a}^b kx^{a+b-k} +[ (1+b)x^a-ax^{b+1}]\]\[\large x \sum_{k=a}^{b}k x^{a+b-k}- \sum_{k=a}^b kx^{a+b-k} +[ (1+b)x^a-ax^{b+1}]\] \[\large (x -1) \sum_{k=a}^b kx^{a+b-k} +[ (1+b)x^a-ax^{b+1}]\]

To recover @Marki 's formula, just let x=10, a=0, and b=n and everything is exactly the same =D

Answer 45

looks neat xD i hope we can do things like changing the base etc and have a little more fun
enjoy your breakfast @Kainui :)

Answer 46

xD give me two weeks only and i promise to solve everything xD

Answer 47

prepare for ur exams
@Kainui please ban Marki

Answer 48

bhahaha he wont dare :P

Answer 49

ok, enjoy ur bf kai xD

Answer 50

i'll go now .

Answer 51

Hahaha XD Yeah I wouldn't ban Marki, she'll find a way to come track me down and make me pay for it. I'll let ganeshie ban her haha XD

Answer 52

@rational We can also see other stuff in base 10 too! Like if you plug in x=10^2 we will get another pattern of things.

Answer 53

Here's one example, and I'll show you another one in a second:

Answer 54

99*(0)+1 = 1
99*(1)+2 = 101
99*(102)+3 = 10101
99*(10203)+4 = 1010101
99*(1020304)+5=101010101
... see! Very fascinating!

Answer 55

hmm yes