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.
\[\sum_{k=0}^{5}\rightarrow \left(\begin{matrix}5 \\ k\end{matrix}\right)\] is what it looks like but with brackets and not perenthesis.
\[\left(\begin{matrix}5 \\ 0\end{matrix}\right)+\left(\begin{matrix}5 \\1\end{matrix}\right)+\left(\begin{matrix}5 \\2\end{matrix}\right)+\left(\begin{matrix}5 \\3\end{matrix}\right)+\left(\begin{matrix}5 \\ 4\end{matrix}\right)+\left(\begin{matrix}5 \\5\end{matrix}\right)\] \[=\left(\begin{matrix}5+5+5+5+5+5 \\0+1+2+3+4+5\end{matrix}\right)=\left(\begin{matrix}30 \\15\end{matrix}\right)\]
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?
is this supposed to be \(\binom{n}{k}\) the binomial aka "n choose k"
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
if so, answer is \(2^n\) or in your case \(2^5\)
\[\sum_{k=0}^n\binom{n}{k}=2^n\] always
you can see this if you add up any row in pascal's triangle, or you can prove it by writing \[2^n=(1+1)^n=\sum_{k=0}^n\binom{n}{k}\]
but if that is not clear, we can just add them one by one and see that you get 32
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!
My teacher weirdly put it on the test and for some reason we never went over an example or anything like it!
do you know how to compute \(\binom{5}{0}\) and \(\binom{5}{1}\) etc?
Yes I believe so. I just never saw it in sigma notation like that.
that just means you are adding them up, add across a row in pascal's triangle
Alright, thanks for the help! :D
yw
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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