Sequence S is defined as follows: S_1 = 2, S_2 = 2^1, S_3 = 2^2, ...

Question

Sequence S is defined as follows:   S_1 = 2 ,  S_2 = 2^1 ,  S_3 = 2^2 , ...,  S_n = 2^{n-1} . What is the sum of the terms is sequence S when  n=10  ?

A.  2^9

B.  2^{10}

C.  2^{16}

D.  2^{35}

E.  2^{37}

CrackverbalGMAT · Answer

Hi Bunuel, wouldn't the first term be  S_1 = 2^0  as  S_n = 2^{n - 1}

The 10 terms are  2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6, 2^7, 2^8 \space and \space 2^9

This is a GP, where the common ratio (r) = 2, the first term (a) = 1 and n = 10

Sum of n terms of a GP, when r > 1 =  \frac{a(r^n - 1)}{r - 1} = \frac{1(2^{10} - 1)}{2 - 1} \approx 2^{10}

Option B

Arun Kumar

chacinluis · Answer

We have the sequence 2, 2^1, 2^2, 2^3, .....2^(n-1)
We want to find the sum of the first ten terms.

Since the first term doesn't come from the form Sn=2^(n-1)

Tenth term is 2^(10-1)=2^9
We'll first find the sum for the terms 2^1, 2^2, ...., 2^9 and then add 2 to it.

Sum of geometric progression with a1=2 and r=2 and n=9

a1(1-r^n)/(1-r) = 2(1-(2^9))/(1-2) = 2(1-2^9)/(-1) = 2(2^9-1) = 2^10 - 2

Then we add the initial 2 to get
2^10 -2+2= 2^10

Final Answer: B

IanStewart · Answer

The question does need to make clear that the rule "s_n = 2^(n-1)" does not apply to the first term s_1, which is defined separately.

If you just look at the sums of the first few terms here, as in most sequence questions, a pattern quickly emerges:

• sum of the first 2 terms is 2 + 2 = 2^2
• sum of the first 3 terms is 2^2 + 2^2 = (2)(2^2) = 2^3
• sum of the first 4 terms is 2^3 + 2^3 = (2)(2^3) = 2^4

and so on, so the sum of the first 10 terms is 2^10.

Sequence S is defined as follows: S_1 = 2, S_2 = 2^1, S_3 = 2^2, ...

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Sequence S is defined as follows: S_1 = 2, S_2 = 2^1, S_3 = 2^2, ...

Prep Toolkit

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