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# Question of the Week - 37 (The nth term of a sequence...)

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Question of the Week - 37 (The nth term of a sequence...)  [#permalink]

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Updated on: 28 Feb 2019, 00:54
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Question Stats:

28% (02:44) correct 73% (02:47) wrong based on 40 sessions

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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. $$\frac{650}{462}$$

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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.
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Re: Question of the Week - 37 (The nth term of a sequence...)  [#permalink]

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22 Feb 2019, 06:18
EgmatQuantExpert wrote:
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. $$\frac{650}{462}$$

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

$$a_1 = \frac{1}{1(1+2)} = \frac{1}{1*3} = \frac{1}{2}[(\frac{1}{1}) - (\frac{1}{3})]$$
$$a_2 = \frac{1}{2(2+2)} = \frac{1}{2*4}= \frac{1}{2}[(\frac{1}{2}) - (\frac{1}{4})]$$
$$a_3 = \frac{1}{3(3+2)} = \frac{1}{3*5}= \frac{1}{2}[(\frac{1}{3}) - (\frac{1}{5})]$$
$$a_4 = \frac{1}{4(4+2)} = \frac{1}{4*6}= \frac{1}{2}[(\frac{1}{4}) - (\frac{1}{6})]$$
$$a_5 = \frac{1}{5(5+2)} = \frac{1}{5*7}= \frac{1}{2}[(\frac{1}{5}) - (\frac{1}{7})]$$

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
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Re: Question of the Week - 37 (The nth term of a sequence...)  [#permalink]

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27 Feb 2019, 05:41

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 Similarly, if we add the next odd term then 1/3 will get cancelled.
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
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Re: Question of the Week - 37 (The nth term of a sequence...)   [#permalink] 27 Feb 2019, 05:41
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# Question of the Week - 37 (The nth term of a sequence...)

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