Can someone explain to me, how this is true?

>>9146237
solve by induction. You can do it in your head.

File: mathandstuff.png (3KB, 252x135px) Image search: [Google]

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Correction

>>9146243
I don't understand how many terms that sum has, could you elaborate?

File: mathandstuff.png (3KB, 259x139px) Image search: [Google]

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JESUS, I'm too stupid for latex.
3 time's a charm.

Its not true, its wrong.

>>9146247
>>9146237

It has (number of last term - number of first term + 1) like all sums.
So (2^{n+1} - 1 - 2^n +1) terms. Which reduce to
(2^{n+1} - 2^n )
2^{n+1} = 2*2^n tho. So if we put 2^n in factor of this sum, we got
2^n*(2-1), which reduce to 2^n.

>>9146247
It has [math]2^{n+1}-1 - 2^n +1[/math] terms.
In other words it has [math]2^n[/math] terms.
You are literally doing 1+1+1+1+1+1... with [math]2^n[/math] 1's.
For reference, [math]\sum _{n=a}^b1 = b-a+1[/math].

>>9146255
[eqn]\sum_{k\,=\,2^n}^{2^{n\,+\,1}\,-\,1}1 \,=\, \underbrace{2^{n\,+\,1}\,-\,1\,-\,2^n\,+\,1}_\text{number of terms summed} \,=\, 2^{n\,+\,1}\,-\,2^n \,=\, 2^n\,\left(2\,-\,1\right) \,=\, 2^n[/eqn]

>>9146244
Because, you brainlet, [math] 2^{n+1} = 2 \cdot 2^n[/math]. Chop the first [math]2^n - 1[/math] off, and you're left with [math]2^{n} + 1[/math]. Therefore, we substract one from the entire thing and are left with regular old [math]2^n[/math].