Question

Let ${\sum }_{k=1}^{10}f(a+k)=16(2^{10}-1)$, where the function $f$ satisfies $f(x+y)=f\left(x\right)f\left(y\right)$ for all natural numbers $x,y$ and $f\left(1\right)=2$. Then, the natural number ' $a$ ' is:

Answer 1

$\text{Given,}f(x+y)=f\left(x\right)\cdot f\left(y\right)\text{Let}f\left(x\right)={\lambda }^{x}f\left(1\right)=2\therefore \lambda =2$
So, ${\sum }_{k=1}^{10}f(a+k)={\sum }_{k=1}^{10}{\lambda }^{a+k}={\lambda }^0\left({\sum }_{k=1}^{10}{\lambda }^k\right)$
$$=2^a\left[\frac{2(2^{10}-1)}{2-1}\right]2^a[2^1+2^2+2^3+\dots \dots +2^{10}]$$
[by using formula of sum of $n$-terms of a GP having first term ' $a$ ' and common ratio ' $r$ ', is
$$S_n=\frac{a(r^n-1)}{r-1},\text{where}r>1]$$
$$\begin{array}{cc}& \Rightarrow 2^{a+1}(2^{10}-1)=16(2^{10}-1)\text{(given)}\\ & \Rightarrow 2^{a+1}=16=2^4\Rightarrow a+1=4\Rightarrow a=3\end{array}$$