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Mathematics 31 Online
OpenStudy (anonymous):

Evaluate the expression 5 Σ [5] [K] K=0 The 5 is over the K, but there is not a division sign, they are inside 1 pair of brackets together.

OpenStudy (anonymous):

\[\sum_{k=0}^{5}\rightarrow \left(\begin{matrix}5 \\ k\end{matrix}\right)\] is what it looks like but with brackets and not perenthesis.

OpenStudy (anonymous):

\[\left(\begin{matrix}5 \\ 0\end{matrix}\right)+\left(\begin{matrix}5 \\1\end{matrix}\right)+\left(\begin{matrix}5 \\2\end{matrix}\right)+\left(\begin{matrix}5 \\3\end{matrix}\right)+\left(\begin{matrix}5 \\ 4\end{matrix}\right)+\left(\begin{matrix}5 \\5\end{matrix}\right)\] \[=\left(\begin{matrix}5+5+5+5+5+5 \\0+1+2+3+4+5\end{matrix}\right)=\left(\begin{matrix}30 \\15\end{matrix}\right)\]

OpenStudy (anonymous):

What is the point of it being on the bottom in the brackets and under it though if it is not division or anything? Can you explain it a bit more?

OpenStudy (anonymous):

is this supposed to be \(\binom{n}{k}\) the binomial aka "n choose k"

OpenStudy (raden):

looks like calculating in combination formula. do you know factorial ? it like : n! = n(n-1)(n-2)(n-3)... 3.2.1 nCr = n!/(r! * (n-r)!) (5 0) --> n = 5 , r = 0 5C0 = 5!/(0! * (5-0)!) = 5! / 0!5! = 1 (5 1) --> n = 5 , r = 1 5C1 = 5!/(1! * (5-1)!) = 5! / 1!4! = 5 ... ... and so on tiil (5 5) continue it

OpenStudy (anonymous):

if so, answer is \(2^n\) or in your case \(2^5\)

OpenStudy (anonymous):

\[\sum_{k=0}^n\binom{n}{k}=2^n\] always

OpenStudy (anonymous):

you can see this if you add up any row in pascal's triangle, or you can prove it by writing \[2^n=(1+1)^n=\sum_{k=0}^n\binom{n}{k}\]

OpenStudy (anonymous):

but if that is not clear, we can just add them one by one and see that you get 32

OpenStudy (anonymous):

Oh thank you for the reply! It's the first time I have seen a problem like this and I'm not used to it. But your explanation helped!

OpenStudy (anonymous):

My teacher weirdly put it on the test and for some reason we never went over an example or anything like it!

OpenStudy (anonymous):

do you know how to compute \(\binom{5}{0}\) and \(\binom{5}{1}\) etc?

OpenStudy (anonymous):

Yes I believe so. I just never saw it in sigma notation like that.

OpenStudy (anonymous):

that just means you are adding them up, add across a row in pascal's triangle

OpenStudy (anonymous):

Alright, thanks for the help! :D

OpenStudy (anonymous):

yw

OpenStudy (anonymous):

if we take it as matrix then i have solved and if it is a binomial co-efficient then it is solved above.

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