Question

If $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,yϵR,f\left(1\right)=2$ and $\sum _{k=1}^{n}f(4+k)=480$, then $n=$

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$4$

On observation, we can say that

$f\left(x\right)=a^{bx}$

$f\left(1\right)=2=a^b$

Now, ${\sum }_{k=1}^{n}f(4+k)=f\left(5\right)+f\left(6\right)+f(7+).....+f(4+n)=480$

so, $a^{5b}+a^{6b}+a^{7}b+....+a^{(4+n)b}=480$

$⟹2^5+2^6+2^7+.....+2^{(4+n)}=480$

Using GP we get

$\frac{2^5(2^{m+4-5+1}-1)}{2-1}=480⟹2^n=16$

So, $n=4$