The weight of a sequence $a_{0},a_{1} , ............. , a_{n-1}$ of real numbers is defined as $a_{0}+\frac{a_{1}}{2} + ..............+ \frac{a_{n-1}}{2^{n-1}}$.
$X$ denote the maximum possible weight of a subsequence of $a_{0},a_{1} , ............. , a_{n-1}$ and
$Y$ the maximum possible weight of a subsequence of $a_{1} , ............. , a_{n-1}$
As the weight of the sequence $a_{0},a_{1} , ............. , a_{n-1}$ of real numbers is defined as $a_{0}+\frac{a_{1}}{2} + ..............+ \frac{a_{n-1}}{2^{n-1}}$. .
so similarly for the subsequence $a_{1} , ............. , a_{n-1}$ the weight can be defined as ($a_{1}+\frac{a_{2}}{2}+.......... + \frac{a_{n-1}}{2^{n-2}}$)…
CASE 1: Now suppose the maximum possible weight of the subsequence is $Y$ only then it will be also maximum weight for the subsequence $a_{0},a_{1} , ............. , a_{n-1}$.
i.e $X=weight(Y)$
CASE 2: if the weight $Y$ is not the maximum then we have to also include $a_{0}$ in it. and as it is given in the question that the maximum weight of the sequence $a_{0},a_{1} , ............. , a_{n-1}$ is $a_{0}+\frac{a_{1}}{2} + ..............+ \frac{a_{n-1}}{2^{n-1}}$. .
but we have the weight $Y$ as ($a_{1}+\frac{a_{2}}{2}+.......... + \frac{a_{n-1}}{2^{n-2}}$)… so before adding $a_{0}$ to it we have to divide this weight by $2$ . so after dividing by $2$ we will get ( $\frac{a_{1}}{2}+ .......... + \frac{a_{n-1}}{2^{n-1}}$).
now after including $a_{0}$ we will get the maximum weight of subsequence $X$ = $a_{0}+\frac{a_{1}}{2} + ..............+ \frac{a_{n-1}}{2^{n-1}}$. where $\frac{a_{1}}{2}+ .......... + \frac{a_{n-1}}{2^{n-1}}$ = $Y$
i.e $X = a_{0} +\frac{Y}{2}$
which is the maximum weight in this case.
so from both CASE 1 and CASE 2 we can conclude that $X$ will be
$X = max( Y, a_{0} + \frac{Y}{2})$
ANS is $option$ $B$ !!