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Series & Sequences

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by [email protected] » Mon Aug 24, 2015 12:05 pm
In the sequence s1, s2, s3,....,sn each term after the first is one more than twice the previous term. If s1=3 then what is the value of (s21-s20) ?
A) 2^19
B) 2^20
C) 2^21
D) (2^20) - 1
E) (2^20) + 1
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by Brent@GMATPrepNow » Mon Aug 24, 2015 12:22 pm
[email protected] wrote:In the sequence s1, s2, s3,....,sn each term after the first is one more than twice the previous term. If s1=3 then what is the value of (s21-s20) ?
A) 2^19
B) 2^20
C) 2^21
D) (2^20) - 1
E) (2^20) + 1
Let's list a few terms of the sequence to get a feel for the pattern:
s1 = 3
s2 = (2)(3) + 1 = 7
s3 = (2)(7) + 1 = 15
s4 = (2)(15) + 1 = 31
s5 = (2)(31) + 1 = 63

See the pattern yet? Here it is:
s1 = 3 = (2^2) - 1
s2 = (2)(3) + 1 = 7 = (2^3) - 1
s3 = (2)(7) + 1 = 15 = (2^4) - 1
s4 = (2)(15) + 1 = 31 = (2^5) - 1
s5 = (2)(31) + 1 = 63 = (2^6) - 1
.
.
.
.
s20 = (2^21) - 1
s21 = (2^22) - 1

So, s21 - s20 = (2^22) - 1 - (2^21) - 1
= (2^22) - (2^21)
= (2^21)(2^1 - 1)
= (2^21)(2 - 1)
= (2^21)(1)
= C

Here's a free video on sequences: https://www.gmatprepnow.com/module/gmat- ... /video/923

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Max@Math Revolution » Mon Aug 24, 2015 12:49 pm
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.



==> S_1=3=(2^2)-1 (Because twice --> 2^n)

S_2=1+2S_1=1+2*3=7=(2^3)-1

S_3=1+2S_2=1+2*7=15=(2^4)-1..........................



S_21=(2^22)-1 and S_20=(2^21)-1,
then S_21-S-20=(2^22)-1-[(2^21)-1]=(2^22)-1-(2^21)-1=(2^22)-(2^21)=(2-1)(2^21)=1*2^21=2^21.



Therefore the best answer is C


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