Problem Solving

A function f(x) defined on the set of whole numbers is such that f(x + y) = f(x) × f(y).

If it is given that f(1) = (1 ÷ 3), find the sum of the following infinite series:

f(0) + f(1) + f(2) + f(3) + ....


1) (1 ÷ 2)
2) (3 ÷ 2)
3) (2 ÷ 3)
4) The given infinite series is not convergent and, hence, the required sum is not a finite number.

OA : 2

Please help...not able to understand this concept..

Hi,
f(x+y) = f(x)*f(y)
Let y = 1
f(x+1) = f(x)*f(1) = (1/3)f(x)
So,f(1) = (1/3)f(0) => 1/3 = (1/3)f(0).
So, f(0) = 1
Every term is (1/3) times the previous term. So, it is an infinite geometric series.
Sum of infinite series with first term 'a' and common ratio 'r' is a/(1-r), where |r| < 1
So, sum of the series is 1/(1-1/3) = 1/(2/3) = 3/2

Hence 2

krishnasty wrote:A function f(x) defined on the set of whole numbers is such that f(x + y) = f(x) × f(y).

If it is given that f(1) = (1 ÷ 3), find the sum of the following infinite series:

f(0) + f(1) + f(2) + f(3) + ....


1) (1 ÷ 2)
2) (3 ÷ 2)
3) (2 ÷ 3)
4) The given infinite series is not convergent and, hence, the required sum is not a finite number.

OA : 2

Please help...not able to understand this concept..

f(1+0)=f(1)*f(0);
also f(1)=1/3;
therefore; 1/3=1/3*f(0);
f(0)=1;
f(0)+f(1)+f(2)+f(3)+........
f(2)=f(1+1)=1/3*1/3=(1/3)^2;
f(3)=f(2+1)=(1/3)^2*1/3=(1/3)^3;
.
.
.

thus, f(0)+f(1)+f(2)+f(3)+........= 1+1/3+1/3^2+1/3^2+...

which is an infinite g.p. whose common ratio is 1/3 and first term is 1,, therefore its sum is a/1-r; where -1<r<1;
therefore, (1)/(1-1/3)=3/2;
hence B

Ian Stewart wrote:This is miles away from being a realistic GMAT question. You will *never* be tested on sums of infinite series on the GMAT, nor do you need to know what it means for a series to be 'convergent'. If the author of your prep material thinks this is a realistic practice question, he or she doesn't understand the test.

thnx a bunch Ian..its a relief to know such questions are not a part of GMAT test.

the series is 1+ 1/3+ 1/3^2 + ....

s = a/(1-r) = 1/1-(1/3) = 3/2.

f(0+1) = f(0)*f(1) = f(0)* 1/3 = 1/3
f(0) = 1 then