anonymous
  • anonymous
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 \(a\), \(b\)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
lol, show that a binomial expansion is a binomial expansion? i assume we have to do induction
anonymous
  • anonymous
I forgot to add this:\[\left(\begin{matrix}n \\ k\end{matrix}\right)=\frac{ n! }{ k!(n-k)! }\]
anonymous
  • anonymous
how would i go about proving this with induction?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
since 0 is a nonnegative integer, so start with n=0 as a basis.
anonymous
  • anonymous
ok let me try with that
anonymous
  • 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
  • amistre64
\[(1+x)^0=\binom 00x^0\]
anonymous
  • anonymous
ok this holds then. sorry i'm sort of slow
amistre64
  • 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
  • amistre64
now that we know its true for n=0, redefine it for some n=t in the setup
anonymous
  • 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
  • amistre64
i had a pretty little latex going, but as for par the google snapped on it
amistre64
  • amistre64
we cant use k again since its already part of the setup, so I picked n=t
amistre64
  • amistre64
multiply both sides by (1+x) so that the left side is (1+x)^t (1+x) = (1+x)^(t+1)
amistre64
  • 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
  • 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
  • amistre64
you havent proved it, youve just made a bold assumption at this point
anonymous
  • anonymous
ok i see. what should my next step be then?
amistre64
  • 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
  • amistre64
sum[0,t] (t k) x^k + sum[1,t+1] (t k-1) x^k
anonymous
  • anonymous
did you make the new k=k+1 in the second term there?
amistre64
  • amistre64
if i latex this thing will crash again ...
anonymous
  • anonymous
it's ok i'm following. thanks
anonymous
  • 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
  • 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
  • amistre64
take out a k=0 from the left, and a k=t+1 from the right ...
amistre64
  • 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
  • amistre64
now we can merge the summation notations and work on them
anonymous
  • anonymous
ok let me try
anonymous
  • 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
  • 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
  • amistre64
good, now simplify those binomials into something useful
amistre64
  • amistre64
good
anonymous
  • 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
  • amistre64
lol, why should I let you be deprived of the joy of discovery :) try them out
anonymous
  • anonymous
i know the 1 can go back but i don't know how to make the \(x^{t+1}\) go back
amistre64
  • amistre64
try the middle for k = t+1, does it produce the x^(t+1)?
anonymous
  • anonymous
yes! lol thanks it's right before my eyes
anonymous
  • anonymous
thank you so much!
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
how do i say that it deduces that a,b relationship though?
amistre64
  • amistre64
you essentially would use that process as the backbone, and replace 1+x with a+b
anonymous
  • anonymous
ok
amistre64
  • amistre64
im not really sure why they dont just have you run the process with a=b from the start
amistre64
  • amistre64
er, a+b
amistre64
  • amistre64
let 1=a, let x=b ...
anonymous
  • anonymous
lol. thank you so much. i have more problems but i'll try them myself first.
anonymous
  • anonymous
thank you so much i panicked when i saw the summation symbol
amistre64
  • amistre64
good luck ;) you might have to finagel it a little to get the final results, but it should go pretty smooth
anonymous
  • 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
  • amistre64
if im around .. prolly got about an hour or so before I have to head home and be bored
anonymous
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.