Question

If the tenth term of the sequence $S=1+5+13+29+.\ldots \ldots .$ is $k,$ then $\frac{k}{500}$ is equal to

Answer 1

$S=1+5+13+29+.\ldots \ldots +t_{9}+t_{10}$
$S=.\ldots \ldots 1+5+13+.\ldots \ldots .+t_{8}+t_{9}+t_{10}$
Subtracting we get $0=1+\left(4 + 8 + 16 + . \ldots \ldots 9 \, t e r m s\right)-t_{10}$
$\Rightarrow t_{10}=1+\left(4 + 8 + 16 + . \ldots . . 9 \, t e r m s\right)=1+4\frac{\left(1 - 2^{9}\right) }{\left(1 - 2\right)}$
$=1+4\left(2^{9} - 1\right)=1+4\left(512 - 1\right)$
$=2045$
$\Rightarrow k=2045\Rightarrow \frac{k}{500}=4.09$