2+2+2^2.... MBA.com CAT

[sarahmailings]

Hi,

I've run into trouble with another MBA.com CAT question.

2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8

a)2^9
b)2^10
c)2^16
d)2^35
e)2^37


Generally, when I run into exponent problems that involve base addition that I can't solve I try and factor. But that didn't get me too far in this problem - maybe I've done something wrong. I eventually guessed D under time constraints, but the correct answer is A.

I couldn't find much in the number properties guide to help me. Most of the similar examples involve adding bases with equivalent exponents. A similar approach might help here, but I'm not sure where to start.

I'm grateful for any suggestions.

Thanks.

[graphica]

It is a bit lengthy but can done.

2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

Rewrite the first 2+2 as 2^2

Therefore

2^2[1+1+2+2^2+2^3+2^4+2^5+2^6] Now making 1+1+2=2^2
2^2[2^2{1+1+2+2^2+2^3+2^4}] Now making 1+1+2=2^2
2^2[2^2{2^2(1+1+2+2^2)}] Now making 1+1+2=2^2
2^2[2^2{2^2(2^2+2^2)}] Now making 1+1+2=2^2
2^2[2^2{2^2(2^2(1+1))}] Now making 1+1=2
Finally it comes to
2^2 X 2^2 X 2^2 X 2^2 X 2

Adding the powers since the base is the same makes it
2^2+2+2+2+1 = 2^9

I hope it helps.

[mschwrtz]

Alternatively you could look for a pattern.

2+2=2^2

2+2+2^2=2^3

so you can see that each term in the expression is equal to the sum of all the previous terms, so that adding each term is doubling the previous sum. Doubling=multiplying by 2=increasing the power by 1.

[atul.prasad]

Lets rephrase the question:

2 + 2 + 2^2 + 2^3... 2^8 as

2 + (sum of first 8 terms of a Geometric progression with factor 2 and the first term also as 2)

Sum of n terms of a GP is a (r^n -1 )/(r-1)
where a is the first term, r is the factor and n is the number of terms
On substituting we get:
2 + 2(2^8-1)/(2-1)
= 2 + 2^9 - 2
= 2 ^ 9
(A is the answer)

[jnelson0612]

atul, once again, excellent work!

Thank you,