|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
21 May 2021, 07:58
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
52% (02:47) correct
48%
(02:38)
wrong
based on 132
sessions
History
Date
Time
Result
Not Attempted Yet
In a sequence \( b_1, b_2, b_3\), . . . in which \(b_1\) is the first term, \(b_1 = 8\), and for all integers \(n > 1, b_n\) = \((−1)^{n−1} \\
(b_{n−1} − 1)\). What is the sum of the first 43 terms of this sequence?
A) 1
B) 13
C) 21
D) 29
E) 43
(b_{n−1} − 1)\). What is the sum of the first 43 terms of this sequence?
A) 1
B) 13
C) 21
D) 29
E) 43
21 May 2021, 19:57
Kudos
Bookmarks
sjuniv32
We can straightaway say that alternate numbers are positive and negative. Also each pair of consecutive number will have a difference of 1.
Let us solve it.
\(b_1 = 8\),
and for all integers \(n > 1, b_n\) = \((−1)^{n−1} (b_{n−1} − 1)\)
\(b_2= (−1)^{2−1} (b_{2−1} − 1)= (−1)^{1} (b_{1} − 1)=-1*(8-1)=-7\)
\(b_3= (−1)^{3−1} (b_{3−1} − 1)= (−1)^{2} (b_{2} − 1)=1*(-7-1)=-8\)
\(b_4= (−1)^{4−1} (b_{4−1} − 1)= (−1)^{3} (b_{3} − 1)=-1*(-8-1)=9\)
\(b_5= (−1)^{5−1} (b_{5−1} − 1)= (−1)^{4} (b_{4} − 1)=1*(9-1)=8\)
The terms will repeat thereafter.
So the sequence is
8, -7, -8, 9, 8.....
So the repeating pattern of 4 number adds up to ( 8 +(-7)+(-8)+9) or 2.
Thus first 10 patterns, equalling first 40 terms, add up to 10*2=20.
The last three terms are 8,-7,-8 or they add up to -7.
Sum of the 43 terms =20+(-7)=13
B
10 Jul 2021, 22:09
Kudos
Bookmarks
First we have to find a pattern from there we generalize the pattern which is the intend
b1=8,
b2=(−1)2−1(b2−1−1)=(−1)1(b1−1)=−1∗(8−1)=−7b2=(−1)2−1(b2−1−1)=(−1)1(b1−1)=−1∗(8−1)=−7
b3=(−1)3−1(b3−1−1)=(−1)2(b2−1)=1∗(−7−1)=−8b3=(−1)3−1(b3−1−1)=(−1)2(b2−1)=1∗(−7−1)=−8
b4=(−1)4−1(b4−1−1)=(−1)3(b3−1)=−1∗(−8−1)=9b4=(−1)4−1(b4−1−1)=(−1)3(b3−1)=−1∗(−8−1)=9
b5=(−1)5−1(b5−1−1)=(−1)4(b4−1)=1∗(9−1)=8b5=(−1)5−1(b5−1−1)=(−1)4(b4−1)=1∗(9−1)=8
Hence 8, -7, -8 , 9 recurs and then the pattern repeats therefore
till 40 the sum is 2*10 = 20 +(-8-7+8) = 13
Hence IMO B
b1=8,
b2=(−1)2−1(b2−1−1)=(−1)1(b1−1)=−1∗(8−1)=−7b2=(−1)2−1(b2−1−1)=(−1)1(b1−1)=−1∗(8−1)=−7
b3=(−1)3−1(b3−1−1)=(−1)2(b2−1)=1∗(−7−1)=−8b3=(−1)3−1(b3−1−1)=(−1)2(b2−1)=1∗(−7−1)=−8
b4=(−1)4−1(b4−1−1)=(−1)3(b3−1)=−1∗(−8−1)=9b4=(−1)4−1(b4−1−1)=(−1)3(b3−1)=−1∗(−8−1)=9
b5=(−1)5−1(b5−1−1)=(−1)4(b4−1)=1∗(9−1)=8b5=(−1)5−1(b5−1−1)=(−1)4(b4−1)=1∗(9−1)=8
Hence 8, -7, -8 , 9 recurs and then the pattern repeats therefore
till 40 the sum is 2*10 = 20 +(-8-7+8) = 13
Hence IMO B
