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Updated on: 28 Feb 2019, 00:54
Originally posted by EgmatQuantExpert on 22 Feb 2019, 04:52.
Last edited by EgmatQuantExpert on 28 Feb 2019, 00:54, edited 1 time in total.
Last edited by EgmatQuantExpert on 28 Feb 2019, 00:54, edited 1 time in total.
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e-GMAT Question of the Week #37
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. \(\frac{300}{440}\)
B. \(\frac{325}{462}\)
C. \(\frac{303}{462}\)
D. \(\frac{375}{450}\)
E. \(\frac{650}{462}\)
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. \(\frac{300}{440}\)
B. \(\frac{325}{462}\)
C. \(\frac{303}{462}\)
D. \(\frac{375}{450}\)
E. \(\frac{650}{462}\)
22 Feb 2019, 06:18
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EgmatQuantExpert
\(a_n = \frac{1}{n(n+2)}\)
\(a_1 = \frac{1}{1(1+2)} = \frac{1}{1*3} = \frac{1}{2}[(\frac{1}{1}) - (\frac{1}{3})]\)
\(a_2 = \frac{1}{2(2+2)} = \frac{1}{2*4}= \frac{1}{2}[(\frac{1}{2}) - (\frac{1}{4})]\)
\(a_3 = \frac{1}{3(3+2)} = \frac{1}{3*5}= \frac{1}{2}[(\frac{1}{3}) - (\frac{1}{5})]\)
\(a_4 = \frac{1}{4(4+2)} = \frac{1}{4*6}= \frac{1}{2}[(\frac{1}{4}) - (\frac{1}{6})]\)
\(a_5 = \frac{1}{5(5+2)} = \frac{1}{5*7}= \frac{1}{2}[(\frac{1}{5}) - (\frac{1}{7})]\)
and so on....
\(a_{20} = \frac{1}{20(20+2)} = \frac{1}{20*22}= \frac{1}{2}[(\frac{1}{20}) - (\frac{1}{22})]\)
Adding them we can see that all terms will be cancelled out except
Sum of 20 terms \(= (\frac{1}{2})[ (\frac{1}{1}) - \frac{1}{3} + (\frac{1}{2}) - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \frac{1}{5} - \frac{1}{7} + ------- +\frac{1}{16} - \frac{1}{18} + \frac{1}{17} - \frac{1}{19} + \frac{1}{18} - \frac{1}{20} + \frac{1}{19} - (\frac{1}{21}) + \frac{1}{20} - (\frac{1}{22}) ]\)
Sum of 20 terms \(= (\frac{1}{2})[ (\frac{1}{1}) + (\frac{1}{2}) - (\frac{1}{21}) - (\frac{1}{22}) ]\)
Sum of 20 terms = \(\frac{325}{462}\)
Answer: Option B
27 Feb 2019, 05:41
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Solution
Given
- • The nth term of a sequence a1, a2, a3, … an is given by an = \(\frac{1}{n(n+2)}\)
To Find
- • The sum of the first 20 terms of the sequence.
Approach & Working
We are given an = \(\frac{1}{n(n+2)}\).
- • Hence, a1 = \(\frac{1}{1(1+2)} =\frac{1}{(1×3)}\)
• Similarly, a2 = \(\frac{1}{(2(2+2))}=\frac{1}{(2×4)}\)
- o a3 = \(\frac{1}{(3×5)}\)
o a4 =\(\frac{1}{(4×6)}\) and so on.
o a19 = \(\frac{1}{(19×21)}\)
o a20 = \(\frac{1}{(20×22)}\)
- • Hence, a1 + a2 + a3 + … +a19 + a20 =\(\frac{1}{(1×3)}\) + \(\frac{1}{(2×4)}\) + \(\frac{1}{(3×5)}\) + \(\frac{1}{(4×6)}\) + ... + \(\frac{1}{(19×21)}\) + \(\frac{1}{(20×22)}\)
Let us assume a1 + a2 + a3 + … a20 = S.
- • S = \(\frac{1}{(1×3)}\) + \(\frac{1}{(2×4)}\) + \(\frac{1}{(3×5)}\) + \(\frac{1}{(4×6)}\) + ... + \(\frac{1}{(19×21)}\) + \(\frac{1}{(20×22)}\)
As the denominator of each fraction is a product of 2 numbers with a constant difference of 2, let us multiply both side by 2 so that we can express the numerator of each fraction in terms of its denominator
Let us multiply by 2 on both the sides.
- • 2S = 2(\(\frac{1}{(1×3)}\) + \(\frac{1}{(2×4)}\) + \(\frac{1}{(3×5)}\) + \(\frac{1}{(4×6)}\) + ... + \(\frac{1}{(19×21)}\) + \(\frac{1}{(20×22)}) 2\)
• 2S = \(\frac{2}{(1×3)}\) + \(\frac{2}{(2×4)}\) + \(\frac{2}{(3×5)}\) + \(\frac{2}{(4×6)}\) + ... + \(\frac{2}{(19×21)}\) + \(\frac{2}{(20×22)}\)
- o If we observe carefully then,
- • \(\frac{2}{(1×3)} = \frac{1}{1}- \frac{1}{3}\)
• \(\frac{2}{(2×4)}= \frac{1}{2}- \frac{1}{4}\)
• \(\frac{2}{(3×5)} = \frac{1}{3} - \frac{1}{5}\)
• \(\frac{2}{(4×6)} = \frac{1}{4} - \frac{1}{6}\)
• Similarly, \(\frac{2}{(19×21)} = \frac{1}{19}- \frac{1}{21}\) and \(\frac{2}{(20×22)} = \frac{1}{20} - \frac{1}{22}\)
- o Now, we can observe that odd or even numbered terms are beginning with the same number that the preceding number is ending with and their sign is opposite.
o So, they will get cancelled when added.
- • For example: \(\frac{2}{(1×3)} = \frac{1}{1} - \frac{1}{3} and \frac{2}{(3×5)} = \frac{1}{3} - \frac{1}{5}\) (odd numbered terms)
- o 1/3 will get cancelled when these two terms are added.
- • And, we will get 1- 1/5 i.e. First number of first term - last number of last term.
o Thus, if we add all the odd terms till a20 then:
- • Sum = First number of first odd term - last number of last odd term.
- • \(\frac{2}{(2×4)} = \frac{1}{2} - \frac{1}{4}\) and \(\frac{2}{(4×6)} = \frac{1}{4} - \frac{1}{6}\) (Even numbered terms)
- o As we did with odd terms, the sum of all the even terms till a20 = First number of first even term - last number of last even term.
Thus, 2S = 2a1 + 2a2 + 2a3 + ... + 2a19+ 2a20
- • Or, 2S = Sum of all odd terms till a20 + Sum of all even terms till a20
- o = (2a1 + 2a3 + ... + 2a19) + (2a2 + 2a4 + ... + 2a20)
o = \([ (\frac{1}{1} - \frac{1}{3})+ ... +(\frac{1}{19} - \frac{1}{21}) ] + [ (\frac{1}{2} - \frac{1}{4})+ … +(\frac{1}{20} - \frac{1}{22}) ]\)
o = \((\frac{1}{1} - \frac{1}{21}) + (\frac{1}{2} - \frac{1}{22})\)
o 2S = \(\frac{20}{21} + \frac{10}{22}\)
o S = \(\frac{10}{21}+ \frac{5}{22}\)
o S = \(\frac{325}{462}\)
Hence, option B is the correct answer.
Correct Answer: Option B
