Answer
(a) We verify:
$a\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right) = n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right) + \left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right)$
(b) We show that
$\left( {1 + x} \right)y' = ay$
(c) We prove that
${T_a}\left( x \right) = {\left( {1 + x} \right)^a}$ for $\left| x \right| \lt 1$.
Work Step by Step
(a) The binomial coefficient is given by
$\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right) = \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n + 1} \right)}}{{n!}}$
$\left( {\begin{array}{*{20}{c}}
a\\
0
\end{array}} \right) = 1$
So,
$\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right) = \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n} \right)}}{{\left( {n + 1} \right)!}}$
Since $\left( {n + 1} \right)! = n!\left( {n + 1} \right)$, we get
$\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right) = \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n} \right)}}{{n!\left( {n + 1} \right)}}$
$\left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right) = \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n} \right)}}{{n!}}$
Write out the last second term on the right-hand side such that
$\left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right) = \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n + 1} \right)\left( {a - n} \right)}}{{n!}}$
Thus,
$\left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right) = \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n + 1} \right)}}{{n!}}\left( {a - n} \right)$
$\left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right) = a\left[ {\dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n + 1} \right)}}{{n!}}} \right] - n\left[ {\dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n + 1} \right)}}{{n!}}} \right]$
Since $\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right) = \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n + 1} \right)}}{{n!}}$, so
$\left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right) = a\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right) - n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right)$
Hence, $a\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right) = n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right) + \left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right)$.
(b) We have from Exercise 88: $y = {T_a}\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$.
$y = {T_a}\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$
$ = 1 + ax + \dfrac{{a\left( {a - 1} \right)}}{{2!}}{x^2} + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)\cdot\cdot\cdot\left( {a - n + 1} \right)}}{{n!}}{x^n} + \cdot\cdot\cdot$
Notice that the initial condition: $y\left( 0 \right) = 1$ is satisfied.
Taking the derivative with respect to $x$ gives
$y' = \mathop \sum \limits_{n = 1}^\infty n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^{n - 1}} = a + a\left( {a - 1} \right)x + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)}}{{2!}}{x^2} + \cdot\cdot\cdot$
Multiply by $\left( {1 + x} \right)$, we get
$\left( {1 + x} \right)y' = \mathop \sum \limits_{n = 1}^\infty n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right)\left( {1 + x} \right){x^{n - 1}}$
$\left( {1 + x} \right)y' = \mathop \sum \limits_{n = 1}^\infty n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^{n - 1}} + \mathop \sum \limits_{n = 1}^\infty n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$
Shifting the index of the first sum on the right-hand side to $n=0$ gives
$\left( {1 + x} \right)y' = \mathop \sum \limits_{n = 0}^\infty \left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right){x^n} + \mathop \sum \limits_{n = 1}^\infty n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$
Notice that we can shift the index of the second sum on the right-hand side to $n=0$ without altering the sum because the $0$-th term is zero. Thus,
$\left( {1 + x} \right)y' = \mathop \sum \limits_{n = 0}^\infty \left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right){x^n} + \mathop \sum \limits_{n = 0}^\infty n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$
$\left( {1 + x} \right)y' = \mathop \sum \limits_{n = 0}^\infty \left[ {\left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right) + n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right)} \right]{x^n}$
Using the result from part (a), that is
$a\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right) = n\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right) + \left( {n + 1} \right)\left( {\begin{array}{*{20}{c}}
a\\
{n + 1}
\end{array}} \right)$
we obtain
$\left( {1 + x} \right)y' = \mathop \sum \limits_{n = 0}^\infty a\left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$
Since $y = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$, so $\left( {1 + x} \right)y' = ay$.
(c) We have from Exercise 88: $y = {T_a}\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$.
Evaluate the derivative of $\dfrac{{{T_a}\left( x \right)}}{{{{\left( {1 + x} \right)}^a}}}$:
$\left( {\dfrac{{{T_a}\left( x \right)}}{{{{\left( {1 + x} \right)}^a}}}} \right)' = \dfrac{{y'{{\left( {1 + x} \right)}^a} - a{{\left( {1 + x} \right)}^{a - 1}}y}}{{{{\left( {1 + x} \right)}^{2a}}}}$
From part (b), we know that $\left( {1 + x} \right)y' = ay$. Thus
$\left( {\dfrac{{{T_a}\left( x \right)}}{{{{\left( {1 + x} \right)}^a}}}} \right)' = \dfrac{{y'{{\left( {1 + x} \right)}^a} - {{\left( {1 + x} \right)}^{a - 1}}\left( {1 + x} \right)y'}}{{{{\left( {1 + x} \right)}^{2a}}}}$
$\left( {\dfrac{{{T_a}\left( x \right)}}{{{{\left( {1 + x} \right)}^a}}}} \right)' = \dfrac{{y'{{\left( {1 + x} \right)}^a} - {{\left( {1 + x} \right)}^a}y'}}{{{{\left( {1 + x} \right)}^{2a}}}} = 0$
This implies that $\dfrac{{{T_a}\left( x \right)}}{{{{\left( {1 + x} \right)}^a}}}$ is a constant.
At $x=0$, we have the initial condition $y\left( 0 \right) = {T_a}\left( 0 \right) = 1$.
Since ${\left( {1 + 0} \right)^a} = 1$, so the ratio $\dfrac{{{T_a}\left( x \right)}}{{{{\left( {1 + x} \right)}^a}}}$ is equal to $1$ at $x=0$.
Since $\dfrac{{{T_a}\left( x \right)}}{{{{\left( {1 + x} \right)}^a}}}$ is a constant, we must have $\dfrac{{{T_a}\left( x \right)}}{{{{\left( {1 + x} \right)}^a}}} = 1$ for all $x$.
Therefore, ${T_a}\left( x \right) = {\left( {1 + x} \right)^a}$.
Since the binomial series is valid for $\left| x \right| \lt 1$, we conclude that ${T_a}\left( x \right) = {\left( {1 + x} \right)^a}$ for $\left| x \right| \lt 1$.
