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Answer :
Answer-
[tex]\boxed{\boxed{f(n)=0.8(f(n-1))+10,\ f(0)=150,\ \ n>0}}[/tex]
Solution-
Here, n represents the number of months and f(n) represents the number of laptops in the store after n months.
As the store had 150 laptops in the month of January or at the beginning.
So [tex]f(0)=150[/tex]
Every month, 20% of the laptops were sold and 10 new laptops were stocked in the store.
As 20% of laptops were sold, so 80% were in the store.
So, after one month total number of laptops in the store,
[tex]\Rightarrow f(1)=0.8(150)+10[/tex]
[tex]\Rightarrow f(1)=0.8(f(0))+10[/tex] [tex](\because f(0)=150)[/tex]
Again after one month total number of laptops in the store,
[tex]\Rightarrow f(2)=0.8\times (0.8(150)+10)+10[/tex]
[tex]\Rightarrow f(2)=0.8(f(1))+10[/tex] [tex](\because f(1)=0.8(150)+10)[/tex]
Analyzing the pattern, the recursive function f(n) will be,
[tex]f(n)=0.8(f(n-1))+10,\ f(0)=150,\ \ n>0[/tex]
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Rewritten by : Barada
Answer:
f(n) = 0.8 × f(n -1) + 10, n > 0
Step-by-step explanation:
Given,
Original number of laptops, f(0) = 150, ( where, f(n) represents the number of laptops after n months )
∵ Every month, 20% of the laptops were sold and 10 new laptops were stocked in the store.
After 1 month,
The number of laptops, f(1) = original laptops - 20% of the original laptops + 10
= 150 - 20% of 150 + 10
= (100 - 20)% of 150 + 10
= 80% of 150 + 10
= 0.8 × f(0) + 10
Similarly, After 2 months,
The number of laptops, f(2) = 0.8 × f(1) + 10,
After 3 months, number of laptops, f(3) = 0.8 × f(2) + 10,
........ so, on
Thus, by following the pattern,
The number of laptops after n months,
f(n) = 0.8 × f(n -1) + 10, n > 0