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The first term in sequence Q equals 1, and for all positive integers
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15 Apr 2021, 00:56
15 Apr 2021, 00:56
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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
(A) 91
(B) 127
(C) 216
(D) 343
(E) 784
The first term in sequence Q equals 1, and for all positive integers
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15 Apr 2021, 03:06
15 Apr 2021, 03:06
Quote:
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.
Translate this. Q = 1, |8-1|, |27-8|, ... so on.
Observe two things here. First, the absolute value sign makes no difference because the difference will always be positive.
So you may even imagine it as
Q = 1, 8-1, 27-8, ... so on.
Second, when you sum all the values, pairs such as 1 and -1, and 8 and -8 get cancelled out, leaving only the nth smallest positive perfect cube of the last (or the nth) term in the sum.
Quote:
The sum of the first seven terms in sequence Q is
Now you can imagine that this will only leave out \(7^3\) as it is the last term in the sum.
Hence, D is the correct answer.
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The first term in sequence Q equals 1, and for all positive integers
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08 Jul 2021, 07:09
08 Jul 2021, 07:09
1
The first term in sequence Q equals 1 and
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
This simply means that
\(T_n\) = | \(n^3\) - \((n-1)^3\) |
So, \(T_2\) = | \(2^3\) - \((2-1)^3\) | = | \(2^3\) - \(1^3\) |
As the result is positive so the absolute value sign doesn't really matter here. So we can open the Absolute values to have
\(T_2\) = \(2^3\) - \(1^3\)
Similarly, \(T_3\) = \(3^3\) - \(2^3\) and so on...
\(T_7\) = \(7^3\) - \(6^3\)
Sum of First seven terms is given by
\(T_1\) + \(T_2\) + \(T_3\) + \(T_4\) + \(T_5\) + \(T_6\) + \(T_7\)
= 1 + \(2^3\) - \(1^3\) + \(3^3\) - \(2^3\) + .... + \(7^3\) - \(6^3\)
So, alternate terms will cancel out. Ex \(2^3\) (First part in \(T_2\)) and -\(2^3\)second part in \(T_3\) and so on..
So, we will be left with
= 1 - \(1^3\) + \(7^3\)
= \(7^3\) = 343
So, answer will be D
Hope it helps!
To learn more about Absolute value and Sequences watch the following videos
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
This simply means that
\(T_n\) = | \(n^3\) - \((n-1)^3\) |
So, \(T_2\) = | \(2^3\) - \((2-1)^3\) | = | \(2^3\) - \(1^3\) |
As the result is positive so the absolute value sign doesn't really matter here. So we can open the Absolute values to have
\(T_2\) = \(2^3\) - \(1^3\)
Similarly, \(T_3\) = \(3^3\) - \(2^3\) and so on...
\(T_7\) = \(7^3\) - \(6^3\)
Sum of First seven terms is given by
\(T_1\) + \(T_2\) + \(T_3\) + \(T_4\) + \(T_5\) + \(T_6\) + \(T_7\)
= 1 + \(2^3\) - \(1^3\) + \(3^3\) - \(2^3\) + .... + \(7^3\) - \(6^3\)
So, alternate terms will cancel out. Ex \(2^3\) (First part in \(T_2\)) and -\(2^3\)second part in \(T_3\) and so on..
So, we will be left with
= 1 - \(1^3\) + \(7^3\)
= \(7^3\) = 343
So, answer will be D
Hope it helps!
To learn more about Absolute value and Sequences watch the following videos
