Algebra - 27 (CAT-2007)

A function f(x) satisfies f(1) = 3600, and f(1)+f(2)+........+f(n) = n²f(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) = 2²f(2) = 4f(2) => f(2) = (1/3)*f(1)

For n=3, f(1)+f(2)+f(3) = 3²f(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) = 4²f(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) = 4²f(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)