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

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

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.

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

Kudos, if my explanation was useful for you :D. If not I would love to learn an alternate way!

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

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

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

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

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