The natural number `a` for which `sum_(k=1)^nf(a+k)=16(2^n-1)` where the function `f` satisfies the relation `f (x+ y)= f (x) *f (y) `for all natural numbers x, y and further `f (1) =2` is

Find the natural number a for which sum_(k=1) ^(n)f(a+k)= 16(2^(n)-1) where the function f satisfies the relation f(x+y)=f(x)f(y) for all natural numbers x,y and further f(1)=2 .

Find the natural number a for which sum_(k=1)^(n)f(a+k)=16(2^(n)-1), where the function f satisfies the relation f(x+y)=f(x)f(y) for all natural number x,y and,further,f(1)=2.

The value of natural number 'a' for which sum_(k=1)^(n)f(a+k)=16(2^(n)-1) , where the function satisfies the relation f(x+y)=f(x).f(y) for all natural numbers x, y and further f(1)=2 is

The natural number a for which sum_(k=1)^n f(a+k)=16(2^n-1) where the function f satisfies the relation f(x+y)=f(x).f(y) for all natural numbers x,y and further f(1)=2 is:- A) 2 B) 3 C) 1 D) none of these

Find the natural number a for which sum_(k=1)^nf(a+k)=16(2^n-1), where the function f satisfies the relation f(x+y)=f(x)f(y) for all natural number x , ya n d ,fu r t h e r ,f(1)=2.

Find the natural number a for which sum_(k=1)^nf(a+k)=16(2^n-1), where the function f satisfies the relation f(x+y)=f(x)f(y) for all natural number x , ya n d ,fu r t h e r ,f(1)=2.

Find the natural number a for which sum_(k=1)^nf(a+k)=16(2^n-1), where the function f satisfies the relation f(x+y)=f(x)f(y) for all natural number x , ya n d ,fu r t h e r ,f(1)=2.

Find the natural number a for which sum_(k=1)^nf(a+k)=16(2^n-1), where the function f satisfies the relation f(x+y)=f(x)f(y) for all natural number x , ya n d ,fu r t h e r ,f(1)=2.

Find the natural number a for which sum_(k=1)^nf(a+k)=16(2^n-1), where the function f satisfies the relation f(x+y)=f(x)f(y) for all natural number x , ya n d ,fu r t h e r ,f(1)=2.

The natural number a for which sum_(k=1,n) f(a+k)=16(2^n-1) where the function f satisfies the relation f(x+y)=f(x).f(y) for all natural numbers x,y and further f(1)=2 is:- A) 2 B) 3 C) 1 D) none of these