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A function ƒ(x) satisfies ƒ(1) = 3600 and ƒ(1) + ƒ(2) + … + ƒ(n) = n2 f(n), for all...
A function ƒ(x) satisfies ƒ(1) = 3600 and ƒ(1) + ƒ(2) + … + ƒ(n) = n2 f(n), for all positive integers n > 1. What is the value of ƒ(9)?
A
80
B
240
C
200
D
100
E
120
A
80
B
240
C
200
D
100
E
120
Solution
f(1) + f(2) + f(3) + …. + f(n) = n2f(n)
f(1) = 3600
For n = 2
⇒ f(1) + f(2) = 22f(2)
⇒ f(2) =
For...
f(1) + f(2) + f(3) + …. + f(n) = n2f(n)
f(1) = 3600
For n = 2
⇒ f(1) + f(2) = 22f(2)
⇒ f(2) =
For n = 3
⇒3600{ 1+
} + f(3) = 32f(3)
⇒f(3) = 3600 ×
Similarly
f(9) = 3600 ×
Therefore f(9) = 80
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