## Friday, 16 December 2011

### 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