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03 Dec 2024, 09:00
Kudos
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\(H_{6}\): 85
Flowers: 97
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
70% (03:29) correct
30%
(03:31)
wrong
based on 487
sessions
History
Date
Time
Result
Not Attempted Yet
A species of tree grows in a unique pattern such that its height in a given year follows a sequence defined as:
where \(H_{1}\) = 3 meters and \(H_{2}\) = 5 meters, and \(H_{n}\) is the height of the tree in meters at the end of year n.
The tree produces flowers in odd years (e.g., 1, 3, 5, 7, ...). The number of flowers produced in year 2n+1 is equal to the value of \(H_{2n+1} * n\).
Based on the above information, select for \(H_{6}\) the height of the tree in year 6 and for Flowers the total amount of flowers produced from year 1 to 6.
\(H_{n} = H_{n-1} + 2 * H_{n-2}\)
where \(H_{1}\) = 3 meters and \(H_{2}\) = 5 meters, and \(H_{n}\) is the height of the tree in meters at the end of year n.
The tree produces flowers in odd years (e.g., 1, 3, 5, 7, ...). The number of flowers produced in year 2n+1 is equal to the value of \(H_{2n+1} * n\).
Based on the above information, select for \(H_{6}\) the height of the tree in year 6 and for Flowers the total amount of flowers produced from year 1 to 6.
| \(H_{6}\) | Flowers | |
| 67 | ||
| 79 | ||
| 85 | ||
| 86 | ||
| 97 | ||
| 100 |
ShowHide Answer
Official Answer
\(H_{6}\): 85
Flowers: 97
03 Dec 2024, 12:11
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Can someone explain how they got 97 flowers
03 Dec 2024, 14:19
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cdrectenwald
Flowers are produced only in odd years (n=1,3,5...). For odd year 2n+1, the number of flowers is:
Flowers in year 2n+1 = H2n+1 * n
1. Year 1 (2n+1=1, solving n we get n=0):
Flowers = H1*0 = 3*0 = 0
2. Year 3 (2n+1=3, solving n we get n=1):
Flowers = H3*1 = 11*1 = 11
3. Year 5 (2n+1=5, solving n we get n=2):
Flowers = H5*2 = 43*2 = 86
Total Flowers = 0+11+86 = 97.
