Question

Did you guys notice the 11^n where n is a nonnegative integer digits occur in the pascal's triangle? Just sharing this because I have just noticed this recently and thought it would be cool to share.(note:I'm not claiming I was the first to notice it)
Also I will multiply 11^n with/o doing any carrying..
11^0=1
11^1=11
11^2=121
11^3=1331
11^4=14641
11^5=15(10)(10)51<-this part and parts after this is where I'm talking about I will not carry over
for example 11^4=14641
11^5=14641*11
=14641+146410
= 15(10)(10)51

Then 11^6=11^5*11
=15(10)(10)51+15(10)(10)510
=16(15)(20)(15)61

Answer 1

cool :)
that means
11^n=k(k+1)/2

Answer 2

It'd be interesting to see a combinatoric explanation for this

Answer 3

No doubt something to do with counting the numbers in certain digits places.

Answer 4

11=10+1
wow !
then using binomial !
i really like it

Answer 5

BINGO!

\[11^n = (10+1)^n = \binom{10}{0}10^n + \binom{10}{1}10^{n-1} + \cdots + \binom{10}{n}10^0 \]

cute :D

Answer 6

yep very cute

Answer 7

*

\[11^n = (10+1)^n = \binom{n}{0}10^n + \binom{n}{1}10^{n-1} + \cdots + \binom{n}{n}10^0 \]

Answer 8

increasing powers of 10 account for left shifting so that the result manifests as a row in pascal's triangle !

Answer 9

binomial theorm is so cool :3
great power

Answer 10

Indeed it is! my all time favorite expression involving binomial theorem

\[\large 1+2+3+\cdots n+1 = \binom{n}{2}\]

Answer 11

gauss fan

Answer 12

Another fun fact, the sum of all the digits in 11^n is 2^n. Interesting stuff, I will think about this thanks @myininaya !

Also, I think @Miracrown might like to see what you've found.

Answer 13

I'm lost in the middle of nowhere.

Answer 14

I just noticed, we can find higher orders of pascal's triangle in similar things, so for instance

(a+b+c)^n

you have a tetrahedral pyramid thing so you can sort of see how each corner of the bottom corner is a^3, b^3, or c^3 and as you move from one corner to another you lose one and gain another.

Now if you add up all the same numbers in the column, they exactly give you the numbers in 111^n like I suspected!