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805+ (Hard)|   Sequences|      
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Bunuel
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The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 23 Nov 2023, 02:30
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The \(n_{th}\) term of a sequence \(a_1, \ a_2, \ a_3 \ … \ a_n\) is given by \(a_n = \frac{1}{n(n+2)}\). What is the sum of the first 20 terms of the sequence?

(A) 300/440

(B) 325/462

(C) 303/462

(D) 375/450

(E) 650/462

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B
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The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 24 Nov 2023, 10:34
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Tn = \(\frac{1}{n(n+2)}\) = \(\frac{[m][fraction]1}{n} \)- \(\frac{1}{(n+2)}\)/2[/fraction][/m]

2Sn = \(\frac{1}{1.3}\) + \(\frac{1}{2.4}\)+ \(\frac{1}{3.5}\)...........................\(\frac{1}{18.20}\) + \(\frac{1}{19.21}\) + \(\frac{1}{20.22}\)

2Sn = \(\frac{1}{1}\) - \(\frac{1}{3}\) + \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) ........................\(\frac{1}{18}\) - \(\frac{1}{20}\) + \(\frac{1}{19}\)- \(\frac{1}{21}\) + \(\frac{1}{20}\) - \(\frac{1}{22}\)

2Sn = 1 + \(\frac{1}{2}\) - \(\frac{1}{21}\) - \(\frac{1}{22}\) = \(\frac{650}{462}\)

Sn = \(\frac{325}{462}\)

O\(\)ption B
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The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 28 Nov 2023, 07:51
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sj296
Tn = \(\frac{1}{n(n+2)}\) = \(\frac{[m][fraction]1}{n} \)- \(\frac{1}{(n+2)}\)/2[/fraction][/m]

2Sn = \(\frac{1}{1.3}\) + \(\frac{1}{2.4}\)+ \(\frac{1}{3.5}\)...........................\(\frac{1}{18.20}\) + \(\frac{1}{19.21}\) + \(\frac{1}{20.22}\)

2Sn = \(\frac{1}{1}\) - \(\frac{1}{3}\) + \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) ........................\(\frac{1}{18}\) - \(\frac{1}{20}\) + \(\frac{1}{19}\)- \(\frac{1}{21}\) + \(\frac{1}{20}\) - \(\frac{1}{22}\)

2Sn = 1 + \(\frac{1}{2}\) - \(\frac{1}{21}\) - \(\frac{1}{22}\) = \(\frac{650}{462}\)

Sn = \(\frac{325}{462}\)

O\(\)ption B
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Hey, can you help me understand how you moved to 2Sn and also the first line is a bit confusing. Would really appreciate your help on this. TIA
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The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 28 Nov 2023, 08:25
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The general term of the series has been given as an: \(\frac{1}{(n(n+2))}\)

Now, this can be re-written as 2an = \(\frac{1}{n}\) - \(\frac{1}{(n+2)}\))....................(1)

The sum of n terms Sn = a1 + a2 .....+an......these terms can be written as per the form in equation (1)

So, if you take the 2 in the denominator to LHS, Sn becomes 2Sn.

Hope, I am able to address your queries.

abishekcmouli
sj296
Tn = \(\frac{1}{n(n+2)}\) = \(\frac{[m][fraction]1}{n} \)- \(\frac{1}{(n+2)}\)/2[/fraction][/m]

2Sn = \(\frac{1}{1.3}\) + \(\frac{1}{2.4}\)+ \(\frac{1}{3.5}\)...........................\(\frac{1}{18.20}\) + \(\frac{1}{19.21}\) + \(\frac{1}{20.22}\)

2Sn = \(\frac{1}{1}\) - \(\frac{1}{3}\) + \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) ........................\(\frac{1}{18}\) - \(\frac{1}{20}\) + \(\frac{1}{19}\)- \(\frac{1}{21}\) + \(\frac{1}{20}\) - \(\frac{1}{22}\)

2Sn = 1 + \(\frac{1}{2}\) - \(\frac{1}{21}\) - \(\frac{1}{22}\) = \(\frac{650}{462}\)

Sn = \(\frac{325}{462}\)

O\(\)ption B

Hey, can you help me understand how you moved to 2Sn and also the first line is a bit confusing. Would really appreciate your help on this. TIA
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The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 21 May 2026, 09:15
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The \(n_{th}\) term of a sequence \(a_1, \ a_2, \ a_3 \ ... \ a_n\) is given by \(a_n = \frac{1}{n(n+2)}\).

What is the sum of the first 20 terms of the sequence?

\(a_n = \frac{1}{n(n+2)} = \frac{(n+2)-n}{2n(n+2)} = \frac{1}{2n} - \frac{1}{2(n+2)} = \frac{1}{2} * (\frac{1}{n} - \frac{1}{n+2})\)

\(a_1 + a_2 + ... + a_n = \frac{1}{2} * (\frac{1}{1} - \frac{1}{3 }+ \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + ..... + \frac{1}{19} - \frac{1}{21} + \frac{1}{20} - \frac{1}{22}) = \frac{1}{2} * (\frac{1}{1} + \frac{1}{2} - \frac{1}{21 }- \frac{1}{22}) = \frac{1}{2} * (\frac{3}{2} - \frac{1}{21} - \frac{1}{22}) = \frac{1}{2} * \frac{693-22-21}{462} = \frac{650}{462*2 }= \frac{325}{462}\)

IMO B
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The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 21 May 2026, 12:14
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I tried it manually. Found the first and the last terms and then used the formula (first term+last term)/2 * Sum of terms.
So, (1/3 + 1/440)/2 * 20.

Why is this approach wrong? It was more intuitive for me. Please help Bunuel, KarishmaB bb Kinshook
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The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 21 May 2026, 20:33
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There are many approaches to solve a problem. In most of the cases, there will always be a simpler shorter way. I don't think anyone mentioned that you were wrong. Am I wrong?
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I tried it manually. Found the first and the last terms and then used the formula (first term+last term)/2 * Sum of terms.
So, (1/3 + 1/440)/2 * 20.

Why is this approach wrong? It was more intuitive for me. Please help Bunuel, KarishmaB bb Kinshook
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The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 21 May 2026, 22:55
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I think he/she means that his/her approach did not produce the correct answer . The formula you mentioned applies only to arithmetic sequences,

These:

3 + 5 + 7 + 9 + 11

4.5 + 5 + 5.5 +6 + 6.5 + 7

But not these:

1 + 4 + 9 + 16 + 25

2 + 6 + 18 + 54

1/2. + 1/3 + 1/4
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The nth term of a sequence a1, a2, a3 an is given by an = 1/(n(n+2)) 22 May 2026, 02:58
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I tried it manually. Found the first and the last terms and then used the formula (first term+last term)/2 * Sum of terms.
So, (1/3 + 1/440)/2 * 20.

Why is this approach wrong? It was more intuitive for me. Please help Bunuel, KarishmaB bb Kinshook
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The formula is (First Term + Last Term)/2 * number of terms
and it is used for Arithmetic Progressions only.
The given sequence is not an arithmetic progression so we cannot use this formula here. I suggest you to read up on Arithmetic progression to understand why the formula is what it is.
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