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/sci/ - Science & Math

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5180878 No.5180878 [Reply] [Original]

Prove equivalence:

2^(n-1) = 2^n-2^(n-1)

>> No.5180905

0/10

>> No.5180911

>>5180905
That's a fraction not a proof.

>> No.5180929

>>5180911
Try harder.

>> No.5180970

2^n - 2^(n-1) = 2 * 2^(n-1) - 2^(n-1) = (2 - 1) 2^(n-1) = 2^(n-1)

>> No.5180984

>>5180970
Nice, tnx

>> No.5181027

>>5180878

they don't look equivalent to me.

is it base 10?

>> No.5181053
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5181053

>>5180878
0/10

>> No.5181140

2^(n-1) = 2^n - 2^(n-1)

2^(n-1) + 2^(n-1) = 2*2^(n-1)

2^1*2^(n-1) = 2^(n-1+1) = 2^n

>> No.5181156

>>5180878
Let x = 2^(n-1).

x = 2^n - x
x - x = 2^n
2^n = 0

Where is your god now?

>> No.5181200

>>5181156

>x - x = 2^n
>not x + x = 2^n

disgusting engineer, go back to /fa/

>> No.5181220 [DELETED] 
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5181220

cure result actually. here a more general form

http://www.wolframalpha.com/input/?i=a^n+%3D%3D+a^%28n-m%29+%2B+%28a^m-1%29+a^%28n-m%29

>> No.5181221 [DELETED] 

cute*

>> No.5181222

>>5181156
>>5181156
>>5181156

>"where is your god now?"
>Can't perform simple addition.

>> No.5181234 [DELETED] 

here a more general form of formally breaking up the power to n

<span class="math">a^n == a^(n-m) + (a^m-1) a^(n-m)[/spoiler]

here

a=2
m=1

>> No.5181240

here a more general form of formally breaking up the power to n

<span class="math">a^n = a^{n-m} + (a^m-1) a^{n-m}[/spoiler]

here

a=2
m=1