Mohamed decided to track the number of leaves on the tree in his backyard each year. The first year, there were 500500500 leaves. Each year thereafter, the number of leaves was 40\%40%40, percent more than the year before. Let f(n)f(n)f, left parenthesis, n, right parenthesis be the number of leaves on the tree in Mohamed's back yard in the n^\text{th}n th n, start superscript, start text, t, h, end text, end superscript year since he started tracking it. fff is a sequence. What kind of sequen

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Question:

Mohamed decided to track the number of leaves on the tree in his backyard each year. The first year, there were 500500500 leaves. Each year thereafter, the number of leaves was 40\%40%40, percent more than the year before. Let f(n)f(n)f, left parenthesis, n, right parenthesis be the number of leaves on the tree in Mohamed's back yard in the n^\text{th}n th n, start superscript, start text, t, h, end text, end superscript year since he started tracking it. fff is a sequence. What kind of sequence is it?

Answers

Answer:[tex]f(n) = (1.4)^{(n - 1)} \times 500[/tex]The sequence is a geometric sequence.Step-by-step explanation:The number of leaves in the tree in the first year is 500.And the number of leaves in the tree in the second year increases by 40% of the number in the first year.So, it will be [tex]500(1 + \frac{40}{100}) = 1.4 \times 500[/tex]Again, the number of leaves in the tree in the third year increases by 40% of the number in the second year and so on.Therefore, the number of leaves in the tree in the third year will be [tex]1.4 \times 500 \times (1 + \frac{40}{100}) = (1.4)^{2} \times 500[/tex] and so on.If we want to calculate the number of leaves in the tree in the nth year by f(n) then it will be given by [tex]f(n) = (1.4)^{(n - 1)} \times 500[/tex] Therefore, the sequence is a geometric sequence. (Answer)

Answer:Geometric sequencef(1)=      500 f(n)=f(n-1)      x  1.4Step-by-step explanation:

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