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derive an explicit formula for this sum
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\[\sum_{i=0}^{n}(n ^{n-i} 4^{i})\]
@ganeshie8 @waterineyes
so hard T_T
Is this not the explicit formula itself?
derive the formula without sum symbol
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like find f(n) that represent this sum
put the values of \(i\) one by one: \[\sum_{i=0}^{n}(n ^{n-i} 4^{i}) = 4^0 n^{n-0} + 4^1 n^{n-1} + 4^2 n^{n-2} + .... + 4^n\]
yea, but how do you find an equation, that you can just plug in n, and gives you the answer @_@
So, basically we have this.. :)
Have patience, I also don't know.. :P
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fak :x
OMG I KNOW
Using Binomial?
it's equal to the sum of \[2^{n-i}*2^{2i} = 2^{n+i}\]
really? but how?
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@ganeshie8
you just transform 4^i into 2^2i :P
ok i got the answer :D
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