The first term in sequence Q equals 1, and for all positive integers

Question

The first term in sequence Q equals 1, and for all positive integers n equal to or greater than 2, the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube. The sum of the first seven terms in sequence Q is

(A) 91
(B) 127
(C) 216
(D) 343
(E) 784

Kudos for a correct solution.

(A) 91
(B) 127
(C) 216
(D) 343
(E) 784

dagoberto · Accepted Answer

note: This is my first post for GMAT club so please forgive me if my notation is weird (or non existent)...

we have that Q1=1 (by definition)

for Qn, where n=>2, the formula is Qn= | Q(nth smallest perfect cube) - Q(second smalls perfect cube) |

in order to make things a little easier... lets find some  positive  perfect cubes... which are: 
1^3 : 1 =1 × 1 × 1 (smallest perfect cube)
2^3: 8 =2 × 2 × 2 (2nd smallest perfect cube)
3^3: 27 =3 × 3 × 3
4^3: 64 =4 × 4 × 4
5^3: 125 =5 × 5 × 5
6^3: 216 =6 × 6 × 6
7^3: 343 =7 × 7 × 7
8^3: 512 =8 × 8 × 8
9^3: 729 =9 × 9 × 9
10^3 : 1000 =10 × 10 × 10 (10th smallest perfect cube)

so to complete the Q2, we need: | Q(2nd smallest perfect cube) - Q(1st smallest cube) | = | 8 - 1 | = 7
Q3: | Q(3rd smallest perfect cube) - Q(3rd smallest perfect cube) = | 27 - 8 | = 19....
.
.
.
Q7: |Q(7th smallest perfect cube) - Q(6th smallest perfect cube) = |343-216 | =  127

which would mean answer b? no that's a trap answer... that's the 7th term of the sequence.. not the  sum  of the 1st seven terms...

you can get all the terms and add them if you'd like.. but to quote Sweet Brown "aint nobody got time for that".....

instead lets do the sum of the sequences to see of there is a pattern....

Q1 =1..... which happens to be 1x1x1...
Q1 + Q2 = 1 + 7 = 8 .... happens to be 2x2x2
Q1 + Q2 + Q3 = 1 +7 +19= 27... happens to be 3x3x3....

imma guess that 
Q1 + ....... + Q7= whatever to be 7x7x7.... which is 343.....=>  D

AbhimanyuDhar · Answer

The first term in sequence Q equals 1, and for all positive integers n equal to or greater than 2, the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube. The sum of the first seven terms in sequence Q is
1 terms is 1
2 term will be ( 2^3 - 1^3),
3 term will be ( 3^3 - 2^3),
4 term will be ( 4^3 - 3^3),
5 term will be ( 5^3 - 4^3),
6 term will be ( 6^3 - 5^3),
7 term will be ( 7^3 - 6^3).

if we add all the terms = ( 7^3 - 6^3 +6^3 - 5^3+5^3 - 4^3+ 4^3 - 3^3+ 3^3 - 2^3+ 2^3 - 1^3 +1) we are left with 7 ^3

i have not taken absolute value because we have ( grater positive value)^3 - ( smaller postie value )^3 which will anyway give postie value.

anudeep133 · Answer

Solution :   The question means  Q_n = n^3 - (n-1)^3  for n>=2 and  Q_1 = 1 
Sum of n terms =  n^{3}  = 7*7*7 = 343

Option, D

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

To solve this problem, we must translate the verbal instructions for the construction of the sequence. The first term is easy:  Q_1 = 1 .

Next, we have this difficult wording: “for all positive integers n equal to or greater than 2, the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube.”

So we’re dealing with all the positive integers beyond 1. Let’s take as an example n = 2. The instructions become these: “the second term equals the absolute value of the difference between the second (nth) smallest positive perfect cube and the first (that is, n-1st) smallest positive perfect cube.”

We know we need to consider the positive perfect cubes in order:
1^3 = 1 = smallest positive perfect cube (or “first smallest”).
2^3 = 8 = second smallest positive perfect cube.

The absolute value of the difference between these cubes is 8 – 1 = 7. Thus  Q_2 = 8 – 1 = 7 .
Likewise,  Q_3 = |3^3 – 2^3| = 27 – 8 = 19 , and so on.
Now, rather than figure out each term of Q separately, then add up, we can save time if we notice that the cumulative sums “telescope” in a simple way. This is what telescoping means:

The sum of  Q_2  and  Q_1 = 7 + 1 = 8 . We can also write (8 – 1) + 1 = 8. Notice how the 1’s cancel.
The sum of  Q_3 ,  Q_2  and  Q_1 = 19 + 7 + 1 = 27 . We can also write (27 – 8 ) + (8 – 1) + 1 = 27. Notice how the 8’s and the 1’s cancel.

At this point, we hopefully notice that the cumulative sum of  Q_1  through  Q_n  is just the nth smallest positive perfect cube.

So the sum of the first seven terms of the sequence is 7^3, which equals 343.

The correct answer is D.

Madhavi1990 · Answer

The first term in sequence Q equals 1, and for all positive integers n equal to or greater than 2, the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube. The sum of the first seven terms in sequence Q is

(A) 91
(B) 127
(C) 216
(D) 343
(E) 784

Method I used:

A1= 1 --> 1^3= 1 Thus 1 is smallest cube

Next smallest cube =8 (it is 2^3).

Now the series can be endless, and the pattern of nth term will remain the same.

So A(n) =|1 - 8| = |-7| = difference of 7 {likely to repeat itself}
So, the difference must a power of 7. Which 343.
Ans D

For those extremely sleepy and not paying attention the answer as multiple of 9 was also present to confuse (216). But either way, great question. Took just about 2 minutes when I got it

Kinshook · Answer

T_1 = 1 
 T_2 = 2^3 - 1^3 
 T_3 = 3^3 - 2^3 
 T_4 = 4^3 - 3^3 
 T_5 = 5^3 - 4^3 
 T_6 = 6^3 - 5^3 
 T_7 = 7^3 - 6^3

Sum of first seven terms =  T_1 + T_2 + T_3 + T_4 + T_5 + T_6 + T_7 = 7^3 = 343  since other terms cancel out.
In general  S_n = n^3  for all n>0

IMO D

ScottTargetTestPrep · Answer

We see that the nth term of sequence Q, when n ≥ 2, is:

a_n = n^3 - (n - 1)^3

Writing the second to seventh terms, we have:

a_2 = 8 - 1, a_3 = 27 - 8, a_4 = 64 - 27, a_5 = 125 - 64, a_6 = 216 - 125 and a_7 = 343 - 216.

Since a_1 = 1, the sum of the first seven terms is:

1 + (8 - 1) + (27 - 8) + (64 - 27) + (125 - 64) + (125 - 64) + (216 - 125) + (343 - 216) = 343

Answer: D

Crytiocanalyst · Answer

Key to cracking the question is identfying a pattern and eleminating the terms that recure and find the final value that lasts therefore 
1 terms is 1
2 term will be ( 2^3 - 1^3),
3 term will be ( 3^3 - 2^3),
4 term will be ( 4^3 - 3^3),
5 term will be ( 5^3 - 4^3),
6 term will be ( 6^3 - 5^3),
7 term will be ( 7^3 - 6^3).
Every other term will be cancelled except for 7^3
which is 343 
Hence IMO D

Mehakgyl · Answer

Why did we not consider 0 in this as a part of the nth smallest perfect cube?

Bunuel · Answer

Check the question:

The first term in sequence Q equals 1, and for all positive integers n equal to or greater than 2, the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube. The sum of the first seven terms in sequence Q is

(A) 91
(B) 127
(C) 216
(D) 343
(E) 784

Kudos for a correct solution.

The first term in sequence Q equals 1, and for all positive integers

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

The first term in sequence Q equals 1, and for all positive integers

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards