If n is a positive integer greater than 1, then 2^{n-1} + 2^

Question

If n is a positive integer greater than 1, then  2^{n-1} + 2^n =

A)  3^n 
B)  2^n+1 
C)  2^{2n-1} 
D)  2^{n(n-1)} 
E)  3*2^{n-1}

kusena · Answer

2^n-1+2^n
=>2^n-1(1+2)
=>2^n-1(3)
=>3*2^n-1
Option E

fozzzy · Answer

I didn't understand the manipulation. Could you elaborate...

Bunuel · Answer

If n is a positive integer greater than 1, then 2^{n-1} + 2^n = A) 3^n B) 2^n+1 C) 2^{2n-1} D) 2^{n(n-1)} E) 3*2^{n-1} Factor out 2^{n-1} : 2^{n-1}(1+2)=3*2^{n-1} (notice that 2^{n-1}*2=2^n ). Answer E. Similar questions: if-2-x-2-x-2-3-2-13-what-is-x-127006.html if-5-x-5-x-3-124-5-y-what-is-y-in-terms-of-x-109080.html Hope this helps.

fozzzy · Answer

Here is what I did

2^n . 2^{-1}+ 2^n

we then get  2^n

2^n    --------> I stopped over here

so basically this step is valid----->   2^n . 2^{-1}

Add the exponents we get  2^{n-1}

TeamGMATIFY · Answer

Generally the first thing to do in questions involving exponent is to take out common terms.

2^{n-1} + 2^n =  =  2^{n-1}*(1 + 2)  =  2^{n-1}*3  
Option E

Abhishek009 · Answer

Lets plug in some values of n

given

n = { 2,3,4,5.......}

If n = 2 then ,  2^{2-1} + 2^2  = 6

If n = 3 then ,  2^{3-1} + 2^3  = 12

From n =2 & 3 ; check the result is always a multiple of 3 as well as 2 , now check the options.

Among the given options only option (E) has 3 & 2

If n =2 then ,  3*2^{n-1}  = 6

If n = 3 then ,  3*2^{n-1}  = 12

So, answer is definitely (E)

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If n is a positive integer greater than 1, then 2^{n-1} + 2^

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If n is a positive integer greater than 1, then 2^{n-1} + 2^

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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My Rewards