Getting stuck on 19
#Binomial theroem proof
tight warren
elfin emberBOT
Getting stuck on 19
- Wait patiently for a helper to come along.
- Once someone helps you, say thank you and close the thread with:
+close
- Feel free to nominate the person for helper of the week in #helper-nominations
- Do not ping the mods, unless someone is breaking the rules.
- If you're happy with the help you got here, and the server overall, you can contribute financially as well:
tight warren
The farthest I can get to is equating the coeffecients
sum of (n,k) * (n,k) to n from k = 0 = (2n,n)
lavish river
which step r u asking about?
lavish river
look carefully, u'll notice that this is the coefficient of x^n in (1+x)^2n
this one?
tight warren
yeah like howd u get to this
did u just multiply 2 binomial expansions?
but if thats the case
how come the coeffecients
are different
lavish river
no, wait let me show a more elaborate version of the expansion
lavish river
tight warren
wait can I also view it as this?
lavish river
its doing what a semicolon would do in the following sentence
i have 5 fruits, an apple; a banana; a mango; a berry; and a watermelon
lavish river
well open a few terms of (1+x)^n
then do it again
then multiply the two (1+x)^n expressions
and try to find the coefficient of x^n in (1+x)^n * (1+x)^n
it will be this
lavish river
well if this is the coefficient of x^n in (1+x)^n * (1+x)^n
that means that it is the coefficient of x^n in (1+x)^2n
tight warren
oh I see
but how does that prove the sum of all of the coeffecients
its equal to 2n C n
lavish river
the coefficient of x^n in (1+x)^2n is
tight warren
oh it isnt saying the sum of the coeffecients are 2n C n?
lavish river
u can prove that using the general term of a binomial expansion
uh i dont understand ur question, can u like rephrase it a bit
tight warren
like isnt the question saying if u add up all of the coeffecients of (1+x)^2n
youll get 2n choose n
lavish river
no, the left side of the question isnt the sum of all coefficients, it is the the coefficient of x^n in (1+x)^2n
which will be 2n choose n
tight warren
wait what
sorry for being lost im rlly tired rn lmao
pretty late
and I got school tmr
just been tryna figure out this question
lavish river
np, reread the proof and ask if u got any doubt
lavish river
this is the entire left side right?
tight warren
but how does that prove the sum of all of the coeffecients
this one
well yeah that one too
lavish river
so this is the coefficient of x^n in (1+x)^2n
tight warren
it will be this
how do u find the coeffecient of x^n though did u just manually expand and see it
lavish river
yes i expanded a few terms then saw that this will be the coefficient of x^n
tight warren
yes i expanded a few terms then saw that this will be the coefficient of x^n
can u show me if possible
lavish river
ok i can, but after like 30-40 mins, i got some chem to do rn
tight warren
alr thanks
I just cant visualize it because wouldnt u expand to x^n/2th power for both expansions
then multiply those coeffecients
I dont understand why we're adding them up
lavish river
beacuse different terms will be generated with all these coefficients
so we will take x^n common from them and then add the rest
try finding the coefficient of x^2 in (1+x)^2 * (1+x)^2 using this method
that might help in understanding why we add them
tight warren
try finding the coefficient of x^2 in (1+x)^2 * (1+x)^2 using this method
1
Isn’t that not equal to 2C0 + 2C1 + 2C2?
Or am I losing u
tight warren
Oh 1+x^2 ^2 lmao
I didn’t see the other one
I’m mad tired
Ok yeah
The 2C1 at the end should be 2C0 whoops
Ok
So then off this pattern
U said this is equal to the nth term
Idk if that’s how they want me to prove it though
They gave a hint
It says equate the coefficients
Yeah I think they did it your way
@lavish river how’d you even know to approach it that way
tight warren
Oh
quaint relic
for the last term in the 2nd line shouldn't it be n-n instead of n-1?