show that \[(1+x)^{n}=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)x^{k}\]for any nonnegative integer \(n\) and deduce \[(a+b)^{n}=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)a^{k}b^{n-k}\]for any \(a\), \(b\)

- anonymous

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- amistre64

lol, show that a binomial expansion is a binomial expansion? i assume we have to do induction

- anonymous

I forgot to add this:\[\left(\begin{matrix}n \\ k\end{matrix}\right)=\frac{ n! }{ k!(n-k)! }\]

- anonymous

how would i go about proving this with induction?

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## More answers

- amistre64

since 0 is a nonnegative integer, so start with n=0 as a basis.

- anonymous

ok let me try with that

- anonymous

i'm kind of confused when i try to do the summation of \[\left(\begin{matrix}0 \\ k\end{matrix}\right)\] form 0 to 0?

- amistre64

\[(1+x)^0=\binom 00x^0\]

- anonymous

ok this holds then. sorry i'm sort of slow

- amistre64

if its already know that \[\lim_{x\to -1}(1+x)^0=1\]then we dont have to prove it works for all values of x do we?

- amistre64

now that we know its true for n=0, redefine it for some n=t in the setup

- anonymous

ok and i assume that n=k is true, and then i use n=t+1 to see if it was true, right?

- amistre64

i had a pretty little latex going, but as for par the google snapped on it

- amistre64

we cant use k again since its already part of the setup, so I picked n=t

- amistre64

multiply both sides by (1+x) so that the left side is (1+x)^t (1+x) = (1+x)^(t+1)

- amistre64

we wil need to transform the right side into a t+1 format ... since its the format thats is true for n=0

- anonymous

ok here's what I have:\[(1+x)\sum_{k=0}^{t}\left(\begin{matrix}t \\ k\end{matrix}\right)x^k=\sum_{k=0}^{t+1}\left(\begin{matrix}t+1 \\ k\end{matrix}\right)x^k\]and i don't know if that's right?

- amistre64

you havent proved it, youve just made a bold assumption at this point

- anonymous

ok i see. what should my next step be then?

- amistre64

distribute it thru ....
(1+x) sum[0,t] (t k) x^k
sum[0,t] (t k) x^k + sum[0,t] (t k) x^(k+1)
we need to get both exponents to agree, and we do that by bumping the index by +1

- amistre64

sum[0,t] (t k) x^k + sum[1,t+1] (t k-1) x^k

- anonymous

did you make the new k=k+1 in the second term there?

- amistre64

if i latex this thing will crash again ...

- anonymous

it's ok i'm following. thanks

- anonymous

now i have\[(1+x)^{t+1}=\sum_{k=0}^{t}\left(\begin{matrix}t \\ k\end{matrix}\right)x^k+\sum_{k=1}^{t+1}\left(\begin{matrix}t \\ k-1\end{matrix}\right)x^k\]

- amistre64

\[(1+x)\sum_{k=0}^{t}\binom tk x^k\]
\[\sum_{k=0}^{t}\binom tk x^k+x\sum_{k=0}^{t}\binom tk x^k\]
\[\sum_{k=0}^{t}\binom tk x^k+\sum_{k=0}^{t}\binom tk x^{k+1}\]
\[\sum_{k=0}^{t}\binom tk x^k+\sum_{k=1}^{t+1}\binom t{k-1} x^{k-1+1}\]
\[\sum_{k=0}^{t}\binom tk x^k+\sum_{k=1}^{t+1}\binom t{k-1} x^{k}\]

- amistre64

take out a k=0 from the left, and a k=t+1 from the right ...

- amistre64

\[1+\sum_{k=1}^{t}\binom tk x^k+\sum_{k=1}^{t}\binom t{k-1} x^{k}+\binom t{t+1-1} x^{t+1}\]

- amistre64

now we can merge the summation notations and work on them

- anonymous

ok let me try

- anonymous

I got\[1+\sum_{k=1}^{t}\left[ \left(\begin{matrix}t \\ k\end{matrix}\right) +\left(\begin{matrix}t \\ k-1\end{matrix}\right)\right]x^k+x^{t+1}\]

- anonymous

and i already proved the identity (n,k-1)+(n,k)=(n+1,k) previously, so i can make that (t+1,k).

- amistre64

good, now simplify those binomials into something useful

- amistre64

good

- anonymous

but i'm still left with the 1 and the \(x^{t+1}\), do i just add them back so the summation would be k=0 and t+1?

- amistre64

lol, why should I let you be deprived of the joy of discovery :)
try them out

- anonymous

i know the 1 can go back but i don't know how to make the \(x^{t+1}\) go back

- amistre64

try the middle for k = t+1, does it produce the x^(t+1)?

- anonymous

yes! lol thanks it's right before my eyes

- anonymous

thank you so much!

- amistre64

youre welcome :)

- anonymous

how do i say that it deduces that a,b relationship though?

- amistre64

you essentially would use that process as the backbone, and replace 1+x with a+b

- anonymous

ok

- amistre64

im not really sure why they dont just have you run the process with a=b from the start

- amistre64

er, a+b

- amistre64

let 1=a, let x=b ...

- anonymous

lol. thank you so much. i have more problems but i'll try them myself first.

- anonymous

thank you so much i panicked when i saw the summation symbol

- amistre64

good luck ;) you might have to finagel it a little to get the final results, but it should go pretty smooth

- anonymous

thanks. i have more proof problems. if i don't get it, can i ask for your help again? it's a real analysis problem book.

- amistre64

if im around .. prolly got about an hour or so before I have to head home and be bored

- anonymous

ok i'll open a new problem because i had been thinking about this one for a wekk and can't figure it out

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