Functions 2 Question 18
19. Find the natural number $a$ for which
$$ \sum _{k=1}^{n} f(a+k)=16\left(2^{n}-1\right) $$
where the function $f$ satisfies the relation $f(x+y)=f(x) f(y)$ for all natural numbers $x, y$ and further $f(1)=2$.
(1992, 6M)
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Solution:
- Let $f(n)=2^{n}$ for all positive integers $n$.
Now, for $n=1, f(1)=2=2$ !
$\Rightarrow$ It is true for $n=1$. Again, let $f(k)$ is true.
$$ \begin{array}{rlr} \Rightarrow \quad f(k) & =2^{k} \text {, for some } k \in N . \\ \text { Again, } f(k+1) & =f(k) \cdot f(1) \quad \text { [by definition] } \\ & =2^{k} \cdot 2 \quad \text { [from induction assumption] } \\ & =2^{k+1} \end{array} $$
Therefore, the result is true for $n=k+1$. Hence, by principle of mathematical induction,
$$ f(n)=2^{n}, \forall n \in N $$
Now, $\quad \sum _{k=1}^{n} f(a+k)=\sum _{k=1}^{n} f(a) f(k)=f(a) \sum _{k=1}^{n} 2^{k}$
$$ \begin{aligned} & =f(a) \cdot \frac{2\left(2^{n}-1\right)}{2-1} \\ & =2^{a} \cdot 2\left(2^{n}-1\right)=2^{a+1}\left(2^{n}-1\right) \end{aligned} $$
But $\quad \sum _{k=1}^{n} f(a+k)=16\left(2^{n}-1\right)=2^{4}\left(2^{n}-1\right)$
Therefore, $\quad a+1=4 \Rightarrow a=3$
