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Mathematics 9 Online
OpenStudy (anonymous):

derive an explicit formula for this sum

OpenStudy (anonymous):

\[\sum_{i=0}^{n}(n ^{n-i} 4^{i})\]

OpenStudy (anonymous):

@ganeshie8 @waterineyes

OpenStudy (anonymous):

so hard T_T

OpenStudy (anonymous):

Is this not the explicit formula itself?

OpenStudy (anonymous):

derive the formula without sum symbol

OpenStudy (anonymous):

like find f(n) that represent this sum

OpenStudy (anonymous):

put the values of \(i\) one by one: \[\sum_{i=0}^{n}(n ^{n-i} 4^{i}) = 4^0 n^{n-0} + 4^1 n^{n-1} + 4^2 n^{n-2} + .... + 4^n\]

OpenStudy (anonymous):

yea, but how do you find an equation, that you can just plug in n, and gives you the answer @_@

OpenStudy (anonymous):

So, basically we have this.. :)

OpenStudy (anonymous):

Have patience, I also don't know.. :P

OpenStudy (anonymous):

fak :x

OpenStudy (anonymous):

OMG I KNOW

OpenStudy (anonymous):

Using Binomial?

OpenStudy (anonymous):

it's equal to the sum of \[2^{n-i}*2^{2i} = 2^{n+i}\]

OpenStudy (anonymous):

really? but how?

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

you just transform 4^i into 2^2i :P

OpenStudy (anonymous):

ok i got the answer :D

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