Let a summation expression is given as ${{S}_{n}}=1+q+{{q}^{2}}+........+{{q}^{n}}$ and ${{T}_{n}}=1+\left( \dfrac{q+1}{2} \right)+{{\left( \dfrac{q+1}{2} \right)}^{2}}+.........+{{\left( \dfrac{q+1}{2} \right)}^{n}}$ where q is a real number and $q\ne 1$. If ${}^{101}{{C}_{1}}+{}^{101}{{C}_{2}}{{S}_{1}}+........+{}^{101}{{C}_{101}}{{S}_{100}}=\alpha {{T}_{100}}$, then $\alpha $ is equal to
(A) ${{2}^{100}}$
(B) 200
(C) ${{2}^{99}}$
(D) 202

Hint: We solve this question by first finding the value of ${{S}_{n}}$ and ${{T}_{n}}$ using the formula for sum of n terms of G.P, $\dfrac{{{a}^{n}}-1}{a-1}$. Then we consider the right-hand side of the expression given and simplify it and find its value using the formulas ${}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+......+{}^{n}{{C}_{n}}={{2}^{n}}$ and ${}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}q+{}^{n}{{C}_{2}}{{q}^{2}}+......+{}^{n}{{C}_{n}}{{q}^{n}}={{\left( 1+q \right)}^{n}}$. Then we find the value of the right-hand side of the expression and equate them and solve it to find the value of $\alpha $.

Complete step-by-step solution:
We are given that ${{S}_{n}}=1+q+{{q}^{2}}+........+{{q}^{n}}$.
We are also given that ${{T}_{n}}=1+\left( \dfrac{q+1}{2} \right)+{{\left( \dfrac{q+1}{2} \right)}^{2}}+.........+{{\left( \dfrac{q+1}{2} \right)}^{n}}$.
We can see those it is a sum of $\left( n+1 \right)$ terms that are in G.P with common ratio $q$ and $\left( \dfrac{q+1}{2} \right)$.
Let us consider the formula for the sum of n terms with first term 1 and common ratio $a$ is
$1+a+{{a}^{2}}+......+{{a}^{n-1}}=\dfrac{{{a}^{n}}-1}{a-1}$
Using this formula, we can write ${{S}_{n}}$ as,
$\begin{align}
  & \Rightarrow {{S}_{n}}=1+q+{{q}^{2}}+........+{{q}^{n}} \\
 & \Rightarrow {{S}_{n}}=\dfrac{{{q}^{n+1}}-1}{q-1} \\
\end{align}$
Using the same above formula for sum of n terms we can write ${{T}_{n}}$ as,
\[\begin{align}
  & \Rightarrow {{T}_{n}}=1+\left( \dfrac{q+1}{2} \right)+{{\left( \dfrac{q+1}{2} \right)}^{2}}+.........+{{\left( \dfrac{q+1}{2} \right)}^{n}} \\
 & \Rightarrow {{T}_{n}}=\dfrac{{{\left( \dfrac{q+1}{2} \right)}^{n+1}}-1}{\dfrac{q+1}{2}-1} \\
 & \Rightarrow {{T}_{n}}=\dfrac{{{\left( \dfrac{q+1}{2} \right)}^{n+1}}-1}{\dfrac{q-1}{2}} \\
 & \Rightarrow {{T}_{n}}=\dfrac{{{\left( q+1 \right)}^{n+1}}-{{2}^{n+1}}}{{{2}^{n}}\times \left( \dfrac{q-1}{2} \right)} \\
 & \Rightarrow {{T}_{n}}=\dfrac{{{\left( q+1 \right)}^{n+1}}-{{2}^{n+1}}}{{{2}^{n}}\times \left( q-1 \right)} \\
\end{align}\]
We are given that ${}^{101}{{C}_{1}}+{}^{101}{{C}_{2}}{{S}_{1}}+........+{}^{101}{{C}_{101}}{{S}_{100}}=\alpha {{T}_{100}}..........\left( 1 \right)$.
So, now let us consider ${}^{101}{{C}_{1}}+{}^{101}{{C}_{2}}{{S}_{1}}+........+{}^{101}{{C}_{101}}{{S}_{100}}$. It can also be written as
$\Rightarrow {}^{101}{{C}_{1}}+{}^{101}{{C}_{2}}{{S}_{1}}+........+{}^{101}{{C}_{101}}{{S}_{100}}=\sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}}{{S}_{i}}........\left( 2 \right)$
Now let us substitute the value of ${{S}_{i}}$ in it. Then we get,
$\begin{align}
  & \Rightarrow \sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}{{S}_{i}}}=\sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}\dfrac{{{q}^{i+1}}-1}{q-1}} \\
 & \Rightarrow \sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}{{S}_{i}}}=\dfrac{1}{q-1}\sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}\left( {{q}^{i+1}}-1 \right)} \\
 & \Rightarrow \sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}{{S}_{i}}}=\dfrac{1}{q-1}\sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}{{q}^{i+1}}}-\dfrac{1}{q-1}\sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}} \\
\end{align}$
$\begin{align}
  & \Rightarrow \sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}{{S}_{i}}}=\dfrac{1}{q-1}\left[ {}^{101}{{C}_{1}}q+{}^{101}{{C}_{2}}{{q}^{2}}+......+{}^{101}{{C}_{101}}{{q}^{101}} \right]-\dfrac{1}{q-1}\left[ {}^{101}{{C}_{1}}+{}^{101}{{C}_{2}}+......+{}^{101}{{C}_{101}} \right] \\
 & \Rightarrow \sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}{{S}_{i}}}=\dfrac{1}{q-1}\left[ {}^{101}{{C}_{0}}+{}^{101}{{C}_{1}}q+{}^{101}{{C}_{2}}{{q}^{2}}+......+{}^{101}{{C}_{101}}{{q}^{101}}-{}^{101}{{C}_{0}} \right] \\
 & \text{ }-\dfrac{1}{q-1}\left[ {}^{101}{{C}_{0}}+{}^{101}{{C}_{1}}+{}^{101}{{C}_{2}}+......+{}^{101}{{C}_{101}}-{}^{101}{{C}_{0}} \right] \\
\end{align}$
Now, let us consider the formula for binomial expansion
$\begin{align}
  & \Rightarrow {}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}q+{}^{n}{{C}_{2}}{{q}^{2}}+......+{}^{n}{{C}_{n}}{{q}^{n}}={{\left( 1+q \right)}^{n}} \\
 & \Rightarrow {}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+......+{}^{n}{{C}_{n}}={{2}^{n}} \\
\end{align}$
Using these formulas, we can write the above equation as,
$\begin{align}
  & \Rightarrow \sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}{{S}_{i}}}=\dfrac{1}{q-1}\left[ {{\left( 1+q \right)}^{101}}-1 \right]-\dfrac{1}{q-1}\left[ {{2}^{101}}-1 \right] \\
 & \Rightarrow \sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}{{S}_{i}}}=\dfrac{1}{q-1}\left[ {{\left( 1+q \right)}^{101}}-1-\left( {{2}^{101}}-1 \right) \right] \\
 & \Rightarrow \sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}{{S}_{i}}}=\dfrac{1}{q-1}\left[ {{\left( 1+q \right)}^{101}}-{{2}^{101}} \right] \\
 & \Rightarrow \sum\limits_{i=0}^{100}{{}^{101}{{C}_{i+1}}{{S}_{i}}}=\dfrac{{{\left( 1+q \right)}^{101}}-{{2}^{101}}}{q-1} \\
\end{align}$
Substituting this value in equation (2) we get,
$\Rightarrow {}^{101}{{C}_{1}}+{}^{101}{{C}_{2}}{{S}_{1}}+........+{}^{101}{{C}_{101}}{{S}_{100}}=\dfrac{{{\left( 1+q \right)}^{101}}-{{2}^{101}}}{q-1}..............\left( 3 \right)$
Now let us consider the right-hand side of equation (1). Using the above obtained value of ${{T}_{n}}$, we get
\[\Rightarrow \alpha {{T}_{100}}=\alpha \dfrac{{{\left( q+1 \right)}^{101}}-{{2}^{101}}}{{{2}^{100}}\times \left( q-1 \right)}...........\left( 4 \right)\]
Substituting the value obtained in equations (3) and (4) in equation (1) we get,
$\begin{align}
  & \Rightarrow \dfrac{{{\left( 1+q \right)}^{101}}-{{2}^{101}}}{q-1}=\alpha \dfrac{{{\left( q+1 \right)}^{101}}-{{2}^{101}}}{{{2}^{100}}\times \left( q-1 \right)} \\
 & \Rightarrow 1=\dfrac{\alpha }{{{2}^{100}}} \\
 & \Rightarrow \alpha ={{2}^{100}} \\
\end{align}$
Hence, we get the value of $\alpha $ as ${{2}^{100}}$. Hence, the answer is Option A.

Note: There is a possibility of one making a mistake while solving this problem by writing the value of ${{S}_{n}}$ and ${{T}_{n}}$ as,
$\begin{align}
  & \Rightarrow {{S}_{n}}=1+q+{{q}^{2}}+........+{{q}^{n}} \\
 & \Rightarrow {{S}_{n}}=\dfrac{{{q}^{n}}-1}{q-1} \\
\end{align}$
\[\begin{align}
  & \Rightarrow {{T}_{n}}=1+\left( \dfrac{q+1}{2} \right)+{{\left( \dfrac{q+1}{2} \right)}^{2}}+.........+{{\left( \dfrac{q+1}{2} \right)}^{n}} \\
 & \Rightarrow {{T}_{n}}=\dfrac{{{\left( \dfrac{q+1}{2} \right)}^{n}}-1}{\dfrac{q+1}{2}-1} \\
\end{align}\]
But here there are $\left( n+1 \right)$ terms in their expressions, not $n$ terms so they are wrong.