Consider a function f satisfying f(x + y) = f(x)f(y) where x, y are po

Question

Consider a function f satisfying  f(x + y) = f(x)f(y)  where x, y are positive integers, and  f(1) = 2 . If  f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1)  then  a  is equal to

A. 1
B. 2
C. 3
D. 4
E. 5

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GMATinsight · Answer

f(x + y) = f(x)f(y)

i.e.  f(a + 1) = f(a)f(1) = 2f(a) 
i.e.  f(a + 2) = f  = f(a+1)f(1) = 4f(a) = 2^2f(a) 
i.e.  f(a + 3) = f  = f(a+2)f(1) = 8f(a) = 2^3f(a) 
and so on

f(a + 1) + f(a + 2) + … + f(a + n) = 2f(a)*  = 16(2^n − 1)

Sum of n terms of a Geometric Progression
 a+ar+ar^2+ar^3+---+ar^n = \frac{a(r^n-1)}{(r-1)}

i.e.  2f(a)*  = 2f(a)*

i.e.  2f(a) = 16

i.e.  f(a) = 8

Also,  f(2) = f(1)*(f1) = 2*2 = 4

f(3) = f(2)*(f1) = 4*2 = 8

i.e.  f(3) = f(a)

i.e. a = 3

Answer: Option C

lacktutor · Answer

Consider a function f satisfying  f(x + y) = f(x)f(y)  where x, y are positive integers, and  f(1) = 2 . If  f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1)  then  a  is equal to

Let’s get some values of f(2), f(3),....
 f(1) = 2^{1} 
 f(2) = f( 1+1) = 2*2= 2^{2} 
 f(3) = f( 1+2) = 2*2^{2} = 2^{3} 
.....
 f(n) = ....= 2^{n}

Well,  f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1) 
—>  f(a)f(1) + f(a)f(2) +...+f(a)f(n)  = 
 f(a) 2^1 + f(a) 2^{2} +...+f(a) 2^{n}

—>  f(a) ( 2^{1} + 2^{2} +...+2^{n} ) = 
 f(a) ( 2( 2^n —1) / (2–1) ) = 16 (2^n—1) 
—>  f(a) = 8 
Only f(3) satisfies that equation 
—>  a= 3

Answer (C)

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CareerGeek · Answer

f(x + y) = f(x)f(y) 
-->  f(1 + 1) = f(1)f(1) = 2*2 = 4 = 2^2 
-->  f(3) = f(2 + 1) = f(2)f(1) = 4*2 = 2^3 
So,  f(n) = 2^n

f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1) 
-->  f(a)f(1) + f(a)f(2) + f(a)f(3) + . . . . . . . . f(a)f(n) = 16(2^n − 1) 
-->  f(a)  = 16(2^n − 1) 
--> f(a)*\frac{2(2^n - 1)}{2 - 1} = 16(2^n − 1) 
-->  2^a*2 = 16 
-->  2^{a + 1} = 2^4 
-->  a + 1 = 4 
-->  a = 3

Option C

preetamsaha · Answer

f(x+y)=f(x)f(y)
f(1)=2 = 2^1
putting , x= 1, y=1, 
we get f(1+1)=f(1)f(1)
or, f(2)=2*2 =2^2 
similarly , f(3) = 2^3 , f(4) = 2^4 , ... and so on.

now,f(a+1)+f(a+2)+…+f(a+n)=16(2^n−1)
or, 2^(a+1) + 2^(a+2) +... + 2^(a+n) =16(2^n−1)
or, 2^(a+1)   /(2-1) =16^n−1)  
or, 2^(a+1)   = 2^4(2^n−1)

now, comparing both sides,a+1 = 4 
or, a=3

correct answer will be C

Kinshook · Answer

Asked: Consider a function f satisfying  f(x + y) = f(x)f(y)  where x, y are positive integers, and  f(1) = 2 . If  f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1)  then  a  is equal to

f(x + y) = f(x)f(y) 
Let x = 1 & y =0
f(1) = f(1) f(0) = 2; f(0) =1; f(1) = 2
f(n+1) = f(n)f(1) = 2f(n)
f(n) = 2^n

f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1) 
f(a)(f(1) +f(a)f(2) + ....... + f(a)f(n) = 16(2^n-1)
f(a)(2 + 2^2 +2^3 + ..... 2^n) = 16(2^n-1)
2f(a)(1 + 2 +...... + 2^{n-1}) = 2f(a) (2^n-1) = 16 (2^n-1)
f(a) = 8 = 2^a
Since f(3) = 8
a = 3

IMO C

jatinmittal91 · Answer

Given f(x+y)=f(x)f(y)---------------------------------(eq. 1)
& f(a+1)+f(a+2)+…+f(a+n)=16(2^n−1)-------------(eq. 2)

From Eq. 1 & 2, if n=1
f(a+1)=f(a)f(1)=16(2-1)=16
=> 2f(a)=16
f(a)=8

f(1+1)=f(2)=f(1)f(1)=2x2=4
f(1+2)=f(3)=8=f(a)
Therefore, a = 3

Ans C

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Consider a function f satisfying f(x + y) = f(x)f(y) where x, y are po

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