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805+ Level|   Sequences|      
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Bunuel
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The sequence a1, a2, a3, ..., an, ... is such that an=1/n^21/(n1)^2 Updated on: 21 Aug 2022, 02:17
Originally posted by Bunuel on 26 Jan 2022, 10:00.
Last edited by Bunuel on 21 Aug 2022, 02:17, edited 2 times in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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95% (hard)

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40% (02:31) correct 60% (02:37) wrong based on 83 sessions

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The sequence \(a_1, \ a_2, \ a_3, \ ..., \ a_n, \ ...\) is such that \(a_n=\frac{1}{n^2}−\frac{1}{(n−1)^2}\) for all integers \(n≥2\) and \(a_1 = 1\). Is the sum of this sequence greater than 0.005?

(1) The number of terms in the sequence is greater than 12

(2) The number of terms in the sequence is lesser than 14


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B
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truplayer257
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The sequence a1, a2, a3, ..., an, ... is such that an=1/n^21/(n1)^2 26 Jan 2022, 20:22
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a1 = 1
a2 = 1/(2^2) - 1/(2 - 1)^2 = 1/4 - 1
a3 = 1/(3^2) - 1/4 = 1/9 - 1/4

a1 + a2 + a3 + ....
1 + 1/4 - 1 + 1/9 - 1/4

Do you see the pattern here? Basically, the 2nd term of the expression 1/(n^2) - 1/(n - 1)^2 is cancelling out. So following this logic:

Statement 1:
Since n > 12, the sum has to be less than 1/(12^2) or 1/144. Note that 0.005 is 1/200, so this can or cannot be true if n > 12. Insufficient.

Statement 2:
Since n < 14, and 1/(14^2) = 1/196 > 1/200, the sum of the sequence must be greater than 1/200.
Sufficient.

Answer is B
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The sequence a1, a2, a3, ..., an, ... is such that an=1/n^21/(n1)^2 20 Aug 2022, 11:21
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Hi Bunuel, chetan2u

Can you kindly help me understand why the answer is C and not B ?

thanks
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The sequence a1, a2, a3, ..., an, ... is such that an=1/n^21/(n1)^2 20 Aug 2022, 21:35
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Bunuel
The sequence \(a_1, \ a_2, \ a_3, \ ..., \ a_n, \ ...\) is such that \(a_n=\frac{1}{n^2}−\frac{1}{(n−1)^2}\) for all integers \(n≥2\) and \(a_1 = 1\). Is the sum of this sequence greater than 0.005?

(1) The number of terms in the sequence is greater than 12

(2) The number of terms in the sequence is lesser than 14


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\(a_n= \frac{1}{n^2}−\frac{1}{(n−1)^2}\)...
\(a_1=1\)

\(a_2= \frac{1}{2^2}−\frac{1}{(2−1)^2}\)...
The sum is \(a_1+a_2=1+ \frac{1}{2^2}−\frac{1}{(2−1)^2}=\frac{1}{2^2}\)

\(a_3= \frac{1}{3^2}−\frac{1}{(3−1)^2}\)...
The sum is \(a_1+a_2+a_3=\frac{1}{2^2}+\frac{1}{3^2}−\frac{1}{(3−1)^2}=\frac{1}{3^2}\)...

So, the sum is \(\frac{1}{n^2}\), where n is the number of terms.

Now, \(0.005=\frac{5}{1000}=\frac{1}{200}\) and \(\frac{1}{14^2}>\frac{1}{200}>\frac{1}{15^2}\)
So, we are checking whether the number of terms is less than 15 OR greater than or equal to 15.

(1) The number of terms in the sequence is greater than 12
So, n could be 13 or 14, then the answer would be YES. But 15 and above would give answer as NO.
Insufficient

(2) The number of terms in the sequence is lesser than 14
So, n is always less than 15, and the answer would be YES always.
Sufficient


B

Trent981, I believe OA has been marked C erroneously.
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The sequence a1, a2, a3, ..., an, ... is such that an=1/n^21/(n1)^2 22 Jun 2024, 07:50
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