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Question

Answers

$\left( a \right)\dfrac{{\left( {1000} \right)!}}{{\left( {49} \right)!\left( {951} \right)!}}$

$\left( b \right)\dfrac{{\left( {1001} \right)!}}{{\left( {51} \right)!\left( {950} \right)!}}$

$\left( c \right)\dfrac{{\left( {1000} \right)!}}{{\left( {50} \right)!\left( {950} \right)!}}$

$\left( d \right)\dfrac{{\left( {1001} \right)!}}{{\left( {50} \right)!\left( {951} \right)!}}$

Answer
Verified

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.

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