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# The coefficient of ${x^{50}}$ in the binomial expansion of ${\left( {1 + x} \right)^{1000}} + x{\left( {1 + x} \right)^{999}} + {x^2}{\left( {1 + x} \right)^{998}} + .......... + {x^{1000}}$ is:$\left( a \right)\dfrac{{\left( {1000} \right)!}}{{\left( {49} \right)!\left( {951} \right)!}}$\left( b \right)\dfrac{{\left( {1001} \right)!}}{{\left( {51} \right)!\left( {950} \right)!}}$\left( c \right)\dfrac{{\left( {1000} \right)!}}{{\left( {50} \right)!\left( {950} \right)!}}$$\left( d \right)\dfrac{{\left( {1001} \right)!}}{{\left( {50} \right)!\left( {951} \right)!}}$  Hint: In this particular question first find out that which series is formed by the given series (i.e. whether it is formed A.P, G.P or an H.P series), so after confirming which series is that, find out the number of terms in the series then the sum of the series using standard formulas G.P, so use these concepts to reach the solution of the question.

Given expression
${\left( {1 + x} \right)^{1000}} + x{\left( {1 + x} \right)^{999}} + {x^2}{\left( {1 + x} \right)^{998}} + .......... + {x^{1000}}$
So as we see that the above summation forms a geometric progression summation.
With first term, a = ${\left( {1 + x} \right)^{1000}}$
Common ratio, r = $\dfrac{{x{{\left( {1 + x} \right)}^{999}}}}{{{{\left( {1 + x} \right)}^{1000}}}} = \dfrac{{{x^2}{{\left( {1 + x} \right)}^{998}}}}{{x{{\left( {1 + x} \right)}^{999}}}} = \dfrac{x}{{1 + x}}$
Now as we know that the last term ${a_n}$ of the G.P series is given as,
$\Rightarrow {a_n} = a{r^{n - 1}}$, where n is the number of terms and last term, ${a_n} = {x^{1000}}$
Now substitute the values we have,
$\Rightarrow {x^{1000}} = {\left( {1 + x} \right)^{1000}}{\left( {\dfrac{x}{{1 + x}}} \right)^{n - 1}}$
Now simplify it we have,
$\Rightarrow {\left( {\dfrac{x}{{1 + x}}} \right)^{1000}} = {\left( {\dfrac{x}{{1 + x}}} \right)^{n - 1}}$
Now on comparing we have,
1000 = n – 1
$\Rightarrow n = 1001$
So there are 1001 terms in the given series.
Now as we know the sum of the G.P series is given as,
$\Rightarrow {S_n} = a\dfrac{{\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$, where symbols have their usual meanings.
Now substitute the values in the above equation we have,
$\Rightarrow {S_n} = {\left( {1 + x} \right)^{1000}}\dfrac{{\left( {1 - {{\left( {\dfrac{x}{{1 + x}}} \right)}^{1001}}} \right)}}{{\left( {1 - \dfrac{x}{{1 + x}}} \right)}}$
Now simplify it we have,
$\Rightarrow {S_n} = {\left( {1 + x} \right)^{1000}}\dfrac{{\left( {\dfrac{{{{\left( {1 + x} \right)}^{1001}} - {x^{1001}}}}{{{{\left( {1 + x} \right)}^{1001}}}}} \right)}}{{\left( {\dfrac{1}{{1 + x}}} \right)}}$
$\Rightarrow {S_n} = {\left( {1 + x} \right)^{1001}}\dfrac{{{{\left( {1 + x} \right)}^{1001}} - {x^{1001}}}}{{{{\left( {1 + x} \right)}^{1001}}}}$
$\Rightarrow {S_n} = {\left( {1 + x} \right)^{1001}} - {x^{1001}}$
So this is the sum of the G.P series.
Now as we know that the coefficient of ${x^r}$ term in the Binomial expansion of ${\left( {1 + x} \right)^n}$ is ${}^n{C_r}$.
As, ${\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ..... + {}^n{C_r}{x^r} + ..... + {}^n{C_n}{x^n}$
So the coefficient of ${x^{50}}$ term in the Binomial expansion of ${\left( {1 + x} \right)^{1001}}$ is ${}^{1001}{C_{50}}$.
So the coefficient of ${x^{50}}$ term in the Binomial expansion of (${\left( {1 + x} \right)^{1001}} - {x^{1001}}$) is ${}^{1001}{C_{50}}$.
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property we have,
$\Rightarrow {}^{1001}{C_{50}} = \dfrac{{\left( {1001} \right)!}}{{\left( {50} \right)!\left( {1001 - 50} \right)!}}$
$\Rightarrow {}^{1001}{C_{50}} = \dfrac{{\left( {1001} \right)!}}{{\left( {50} \right)!\left( {951} \right)!}}$
So this is the required answer.
Hence option (D) is the correct answer.

Note:Whenever we face such types of questions the key concept we have to remember always recall the standard formulas of G.P, such as, ${S_n} = a\dfrac{{\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$, where symbols have their usual meanings, and then always recall that the coefficient of ${x^r}$ term in the Binomial expansion of ${\left( {1 + x} \right)^n}$ is ${}^n{C_r}$.

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