Functions 2 Question 2
3.
Let $\sum _{k=1}^{10} f(a+k)=16\left(2^{10}-1\right)$, where the function $f$ satisfies $f(x+y)=f(x) f(y)$ for all natural numbers $x, y$ and $f(1)=2$. Then, the natural number ’ $a$ ’ is
(2019 Main, 9 April I)
(a) 2
(b) 4
(c) 3
(d) 16
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Answer:
Correct Answer: 3. (c)
Solution:
- Given, $f(x+y)=f(x) \cdot f(y)$
Let $\quad f(x)=\lambda^{x}$
[where $\lambda>0$ ]
$\because \quad f(1)=2$
(given)
$\therefore \lambda=2$
So,
$\sum _{k=1}^{10} f(a+k) =\sum _{k=1}^{10} \lambda^{a+k}=\lambda^{a} \sum _{k=1}^{10} \lambda^{k} $
$=2^{a}\left[2^{1}+2^{2}+2^{3}+\ldots . .+2^{10}\right] $
$=2^{a} \frac{2\left(2^{10}-1\right)}{2-1}$
by using formula of sum of $n$-terms of a GP having first term ’ $a$ ’ and common ratio ’ $r$ ‘, is
$ S _n=\frac{a\left(r^{n}-1\right)}{r-1}, \text { where } r>1 $
$ \begin{array}{ll} \Rightarrow & 2^{a+1}\left(2^{10}-1\right)=16\left(2^{10}-1\right) \text { (given) } \\ \Rightarrow & 2^{a+1}=16=2^{4} \Rightarrow a+1=4 \Rightarrow a=3 \end{array} $
