Question

help

Answer 1

noted.

Answer 2

\[\sum_{k=1}^{n}f(a+k)=16(2^n-1)\]

Answer 3

f(x+y)=f(x).f(y)
f(1)=2
find a

Answer 4

i think f(x)=p^x

Answer 5

\[f(a+1) + f(a+2) + \cdots+f(a+n) = 16(2^n-1)\]

\[f(a) [f(1) + f(2) + ... + f(n)] = 16(2^n-1)\]

Answer 6

how did you get f(x) = p^x ?

Answer 7

\[f(a)\sum_{k=1}^{n}f(k)=16(2^n-1)\]

Answer 8

since it satisfies the eqn. and got by observing second condition i gave

Answer 9

wow! f(x) = 2^x seem to work !

Answer 10

a = 4 ?