M08-08

Question

A sequence is defined as follows:
  
 a_1 = 1 
 
 a_2 = -1 
 
 a_{n+1} = a_n + 2a_{n-1} 
  
What is the value of  a_1 + a_2 + ... + a_{1001} ?

A. -2
B. -1
C. 0
D. 1
E. 2

Bunuel · Answer

Official Solution:

A sequence is defined as follows:
  
 a_1 = 1 
 
 a_2 = -1 
 
 a_{n+1} = a_n + 2a_{n-1} 
  
What is the value of  a_1 + a_2 + ... + a_{1001} ?

A. -2
B. -1
C. 0
D. 1
E. 2

Write out more terms of this sequence: 
 
 a_1 = 1 ;
 
 a_2 = -1 ;
 
 a_3 =a_2+2a_1 =-1+2*1=1 ; 
 
 a_4 =a_3+2a_2 =1+2*(-1)=-1 ; 
 
 a_5 =a_4+2a_3 =-1+2*1=1 ; 
 
...
 
Notice that odd terms equal 1 and even terms equal -1. Therefore, the sum  a_1 + a_2 + ... + a_{1001} = 1  (since the even terms cancel out the odd terms up to  a_{1000} , and the addition of  a_{1001} = 1  results in a total of 1).

Answer: D

Senthil7 · Answer

I think this is a  high-quality  question and I agree with explanation.

Vikram_Katti · Answer

I think this is a  high-quality  question and I agree with explanation. This is surely a 600 Lv question

varshaaaaa · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

sssdd1234 · Answer

I think this is a  high-quality  question and I agree with explanation.

Jayant9090 · Answer

I think this is a  high-quality  question and I agree with explanation.

kouassidrissa047 · Answer

I like the solution - it’s helpful.

iBN · Answer

This is a great question that’s helpful for learning.

adityacantstudy · Answer

I did not quite understand the solution. it would be better if you explain in bit detail about the formula uses and increase steps

Bunuel · Answer

We are given
 a_1 = 1 
 a_2 = -1 
 a_{n+1} = a_n + 2a_{n-1}

Now build the next few terms using the formula:
 a_3 = a_2 + 2a_1 = -1 + 2(1) = 1 
 a_4 = a_3 + 2a_2 = 1 + 2(-1) = -1 
 a_5 = a_4 + 2a_3 = -1 + 2(1) = 1 
 a_6 = a_5 + 2a_4 = 1 + 2(-1) = -1

So the pattern is
 1, -1, 1, -1, 1, -1, ...

You can also see why this repeats: whenever two consecutive terms are
 a_{n-1} = 1  and  a_n = -1 ,

then
 a_{n+1} = a_n + 2a_{n-1} = -1 + 2(1) = 1 .

So the next term becomes 1, and the alternation continues.

Now for the sum:

Pair them:
 (a_1 + a_2) + (a_3 + a_4) + ... + (a_{999} + a_{1000}) + a_{1001}

Each pair is
 1 + (-1) = 0

So everything up to  a_{1000}  cancels to 0, and we are left with
 a_{1001} = 1

Therefore,
 a_1 + a_2 + ... + a_{1001} = 1

Answer: D

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