A function f(x) satisfies f(1) = 3600, and
f(1)+f(2)+........+f(n) = n

^{2}f(n), for all positive integers n>1. What is the value of f(9)?
(1)80 (2)240 (3)200 (4)100 (5)120

Solution follows here:

__Solution:__
This is a bit trickier one. We should observe the pattern to
save time.

For n=2, f(1)+f(2) = 2

^{2}f(2) = 4f(2) => f(2) = (1/3)*f(1)
For n=3, f(1)+f(2)+f(3) = 3

^{2}f(3) = 9f(3)
=> 8f(3) = f(1) + (1/3) f(1) = (4/3)f(1)

=> f(3) = (1/6)*f(1)

For n=4, f(1)+f(2)+f(3)+f(4) = 4

^{2}f(4) = 16f(4)
=> 15f(4) = f(1) + (1/3)f(1) + (1/6)f(1) = (3/2)f(1)

=> f(4) = (1/10)*f(1)

For n=5, f(1)+f(2)+f(3)+f(4)+f(5) = 4

^{2}f(5) = 25f(5)
=> 24f(5) = f(1) + (1/3)f(1) + (1/6)f(1) + (1/10)f(1) =
(8/5)f(1)

=> f(5) = (1/15)*f(1)

Observe f(2) = (1/3)f(1), f(3)
= (1/6)f(1), f(4) = (1/10)f(1),

f(5) = (1/15)*f(1)

the multiplicand-fraction follows a series:
1/3,1/6,1/10,1/15...

the denominators are: 3, 3+3, 3+3+4, 3+3+4+5,....

for f(9), the multiplicand-fraction is 1/(3+3+4+5+6+7+8+9) =
1/45

=> f(9) = (1/45)*f(1) = (1/45)*3600 = 80

**Answer (1)**
Answer = (1)80

ReplyDeletef(1) = 3600

3600 + f(2)= 4f(2)-->f(2)= 1200

4800 + f(3)= 9f(3)-->f(3)= 600

5400 + f(4)= 16f(4)-->f(4)= 360

5760 + f(5)= 25f(5)-->f(5)= 240

6000 + f(6)= 36f(6)-->f(6)= 171.428

6171.428 + f(7)= 49f(7)--> f(7)=128.571

6300 + f(8)=64f(8)-->f(8)= 100

6400 + f(9)= 81f(9)-->f(9)= 80

Dear TSK, your answer is correct. But, we need not calculate all values. After finding 3 or 4 a pattern can be observed. That makes the solution much simpler.

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