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Find the natural number "a" for which:-

$$\sum_{k=0}^n f(a+k)=16(2^n-1)$$ where the function $f$ satisfies the relation

$$f(x+y)=f(x)*f(y)$$

for all natural numbers $x,y$ and $f(1)=2$.

I can't figure out how to go about this problem.Any suggestions/solutions would be appreciated.

Omkar
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  • Please note that this property refers to the family of functions $f(n) = c^n$ with c a positive real. $c^{x+y} = c^x \cdot c^y$ – PackSciences May 25 '19 at 13:16
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    Note: $f(2)=f(1)\times f(1)=2\times 2=2^2$. Similarly, $f(3)=f(1+2)=2^3$, and $f(n)=2^n$ for $n\in \mathbb N$. – lulu May 25 '19 at 13:19

2 Answers2

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Note that $f$ is unique, and that $f(n)=2^n$ is a valid $f$. Thus it is the only $f$.

Now, to sum powers of $2$ we use the geometric sum formula to get

$$\sum_{k=0}^n2^{a+k}=2^a\frac{2^{n+1}-1}{2-1}\neq16(2^n-1)$$

for any $k$. Perhaps try double checking the bound of your sum and all off-by-one things that could've gone wrong.

auscrypt
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Firstly, find the function from the second equation. For that, differentiate w.r.t. $y$ (Considering $x$ as constant, you will get

$f'(x+y)=f(x)f'(y)$ and then put $y=0$ which gives $f'(x)=tf(x)$ where $t=f'(0)$ is some constant.

So, from integrating both sides, we get $ln(f(x))=tx+b$ giving $f(x)=ce^{tx}=c(e^t)^x$

Find $c$ after you put $f(x)$ in second equation. $c$ comes out to be $1$

Now use $f(1)=2$. So $f(1)=(e^t)^1=2$ giving $e^t=2$

So, the final function is $f(x)=2^x$

$$2^a\sum_{k=0}^n2^k=2^a(\frac{2^{(n+1)}-1}{2-1})=2^a(2^{n+1}-1)$$

Something is wrong with your original question and match with this expression.

Hope this will be helpful!

Vineet Mangal
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