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Mathematics 21 Online
OpenStudy (anonymous):

Use the binomial theorem to prove that C(n,0)+C(n,1)2+C(n,2)2^2 +⋯+C(n,n)2^n =3^n

OpenStudy (anonymous):

i kn the rows of the pascal triangle is 2^n

OpenStudy (anonymous):

but i thought C(n,n) was just equal to one

OpenStudy (anonymous):

it is

OpenStudy (anonymous):

so how is 2^n=3^n

OpenStudy (anonymous):

only the last term is \(2^n\)

OpenStudy (anonymous):

i have no idea how to do this, but it says "use the binomial theorem" so that has to be a clue

OpenStudy (anonymous):

may be has something to do with \[(1+2)^n=3^n\]?

OpenStudy (anonymous):

that looks promising actually

OpenStudy (anonymous):

thats what am thinking ... but am trying to do it use algebra to get it to it

OpenStudy (anonymous):

expand \[(1+2)^n\] using the binomial theorem and i think you get it

OpenStudy (anonymous):

first term is \[\binom{n}{0}1^n2^0=\binom{n}{0}\] second term is \[\binom{n}{1}1^{n-1}2^1=\binom{n}{1}2\]

OpenStudy (anonymous):

etc it works out to exactly what you want

OpenStudy (anonymous):

that seems to work

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