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

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

\(2+2=2*2=2^2\) (the sum of first and second terms);
\(2^2+2^2=2^3\) (the sum of previous 2 terms and the third term);
\(2^3+2^3=2^4\) (the sum of previous 3 terms and the fourth term);
...
The same will continue and finally we'll get \(2^8+2^8=2*2^8=2^9\), the sum of previous 8 terms and the 9th term.

Answer: A.

OR: we can identify that the terms after the first one represent geometric progression.

Sum of the terms of geometric progression is given by: \(Sum=\frac{a*(r^{n}-1)}{r-1}\), where \(a\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(>1\).

In our original question we have 2 plus G.P. with 8 terms, so:

\(2+(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8)=2+\frac{2*(2^{8}-1)}{2-1}=2^9\).

Answer: A.
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there are nine numbers, so group them up in 3.

\([2+2+2^2]+[2^3+2^4+2^5]+[2^6+2^7+2^8]\)

\([2^3]+2^3[1+2+4]+2^6[1+2+4]=2^3+2^3*7+2^3*8*7\)

\(2^3*64=2^3*2^6=2^9\)

See for pattern:
2 + 2 =2^2
2^2+2^2=2^3
2^3 + 2^3=2^4
........
2^8 + 2^8=2^9

wouldnot take more than 30 seconds.

hope this helps..!!

You can do it using sum of GP as well.

\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = 1 + (1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8)\)
The terms in brackets make a GP with first term = 1 and common ratio = 2

GP = \(1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
Sum of 9 terms of GP \(= 1(1 - 2^9)/(1-2) = 2^9 - 1\)

Sum of the required series = 1 + sum of GP = \(1 + 2^9 - 1 = 2^9\)
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Check similar questions to practice:
3-108690.html
tough-and-tricky-exponents-and-roots-questions-125956-40.html?hilit=geometric#p1029222
sequence-s-is-defined-as-follows-s1-2-s2-2-1-s3-2-2-sn-129435.html
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don't neglect what might be the easiest way to solve this problem -- and what's absolutely the most straightforward: just convert these to concrete numbers.

the problem statement is
2 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256

the first few terms here are pretty insignificant, as far as the total sum is concerned. 32 + 64 is pretty close to 100, so, this sum will be close to 100 + 128 + 256.

the answer choices are
A/ 512
B/ 1024
C/ ha, you must be kidding
D/ ha, you must be kidding
E/ ha, you must be kidding.

this method certainly doesn't take any MORE time than any of the other methods here... and it may in fact take less, for many people.

Posted from my mobile device
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Great explanation kraizada84, one question why is 2^2+2^2=2^3 and not 2^4 ? Same with 2^3 + 2^3=2^4 and not 2^6?....
Thanks for your time in advanced.

Recommend studying up on exponents.

2^2 + 2^2 = 2^2(1+1) = 2^2(2)= 2^3

Exponents are added only when two like terms are multiplied

2^2 * 2^2 = 2^4

Posted from my mobile device

Great thanks Regor60. My silly mistake and make sense now.

Solving using two methods

Method 1: Logic

\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2*2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2*2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= \(2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
Following the same pattern till last term we will get
.
.
.
= \(2^8 + 2^8\)
= \(2*2^8\)
= \(2^9\)

So, Answer will be A

Method 2: Geometric series formula

\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)
= 2 + a Geometric series whose first term is 2, common ratio is 2 and number of terms is 8
= 2 + 2*\(\frac{2^8 - 1 }{ 2 - 1}\)
= 2 + \(2^9\) - 2
= \(2^9\)

So, Answer will be A
Hope it helps!

Watch the following video to learn How to Sequence problems


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