$(a)$$a_{n}=\binom{n}{2}$
Then it will be an infinite series
Series will be $a_{n}=\binom{0}{2},\binom{1}{2},\binom{2}{2},\binom{3}{2},\binom{4}{2},\binom{5}{2}..............$
$a_{n}=0,0,1,3,6,10,15...........$
So, $a_{n}=x^{2}+3x^{3}+6x^{4}+10x^{5}+15x^{6}...........i$
$xa_{n}=$. $x^{3}+3x^{4}+6x^{5}+10x^{6}+15x^{7}...........ii$
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Subtracting i and ii $(1-x)a_{n}=x^{2}+2x^{3}+3x^{4}+4x^{5}+5x^{6}...........$
$=x^{2}(1+2x+3x^{2}+4x^{3}+5x^{4}...........)$
$=\frac{x^{2}}{(1-x)^{2}}$
$a_{n}=\frac{x^{2}}{(1-x)^{3}}$
https://www.quora.com/What-is-the-value-of-coefficient-of-x-5-in-1-2x-3x-2-3-2
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$(b)$Solve from table 1 of book,3rd formula (It is finite series )
$\binom{10}{1},\binom{10}{2},\binom{10}{3},.......$
$=\frac{1}{x}(\binom{10}{1}x+\binom{10}{2}x^{2}+\binom{10}{3}x^{3}+.......)+\frac{1}{x}-\frac{1}{x}$
$=\frac{1}{x}\left ( 1+x \right )^{10}-\frac{1}{x}$