hi, i hv doubt for below question:

find natural number a such that:

sum[f(a+k)] for k=1,n = 16(2^n-1)

where f(x+y)=f(x).f(y) for all natural x,y and f(1)=2.

note:I was able to find out that f(x)=2^x here but not able to proceed further. Please suggest.

thanks

Question

hi, i hv doubt for below question:

find natural number a such that:

sum[f(a+k)] for k=1,n = 16(2^n-1)

where f(x+y)=f(x).f(y) for all natural x,y and f(1)=2.

note:I was able to find out that f(x)=2^x here but not able to proceed further. Please suggest.

thanks

Aman Bansal · Answer

Dear Pawan,

about 80% of the question you have already solved and tha sum can be evaluated using the given condition,

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

sum =

16(2^n-1)

Best Of luck

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hi, i hv doubt for below question: find natural number a such that: sum[f(a+k)] for k=1,n = 16(2^n-1) where f(x+y)=f(x).f(y) for all natural x,y and f(1)=2. note:I was able to find out that f(x)=2^x here but not able to proceed further. Please suggest. thanks

hi, i hv doubt for below question:

find natural number a such that:

sum[f(a+k)] for k=1,n = 16(2^n-1)

where f(x+y)=f(x).f(y) for all natural x,y and f(1)=2.

note:I was able to find out that f(x)=2^x here but not able to proceed further. Please suggest.

thanks

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hi, i hv doubt for below question: find natural number a such that: sum[f(a+k)] for k=1,n = 16(2^n-1) where f(x+y)=f(x).f(y) for all natural x,y and f(1)=2. note:I was able to find out that f(x)=2^x here but not able to proceed further. Please suggest. thanks

hi, i hv doubt for below question:find natural number a such that: sum[f(a+k)] for k=1,n = 16(2^n-1) where f(x+y)=f(x).f(y) for all natural x,y and f(1)=2.note:I was able to find out that f(x)=2^x here but not able to proceed further. Please suggest. thanks

1 Answers

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hi, i hv doubt for below question:

find natural number a such that:

sum[f(a+k)] for k=1,n = 16(2^n-1)

where f(x+y)=f(x).f(y) for all natural x,y and f(1)=2.

note:I was able to find out that f(x)=2^x here but not able to proceed further. Please suggest.

thanks