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03 Dec 2022, 05:51
Kudos
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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)!
Difficulty:
15%
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Question Stats:
81% (01:07) correct
19%
(01:23)
wrong
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If \(s_1, \ s_2, \ s_3, \ ...\) is the sequence such that \(s_n = \frac{n}{n + 1}\) for all positive integers n, then the product of the first 10 terms of this sequence is.
(A) 1/(10*11)
(B) 1/11
(C) 1/10
(D) 9/10
(E) 10/11
(A) 1/(10*11)
(B) 1/11
(C) 1/10
(D) 9/10
(E) 10/11
03 Dec 2022, 08:31
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Bunuel
Let's first get an idea of what the terms look like in the sequence.
\(s_1 = \frac{1}{1 + 1} = \frac{1}{2}\)
\(s_2 = \frac{2}{2 + 1} = \frac{2}{3}\)
\(s_3 = \frac{3}{3 + 1} = \frac{3}{4}\)
\(s_4 = \frac{4}{4 + 1} = \frac{4}{5}\)
At this point we can see the pattern.
So, the product of the first 10 terms of this sequence \(= (\frac{1}{2})(\frac{2}{3})(\frac{3}{4})(\frac{4}{5}).....(\frac{9}{10})(\frac{10}{11})\)
\(= \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10}{ 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11}\)
At this point we can see that several numbers cancel out to get....
\(= \frac{ 1}{11} \)
Answer: B
18 Nov 2024, 15:05
Kudos
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Should the formula be written with a lowercase 's'? I thought the capital 'S' stood for the sum. It's seems misleading.
