Find the coefficient of ${{x}^{50}}$ in the expression ${{\left( 1+x \right)}^{1000}}+2x{{\left( 1+x \right)}^{999}}+3{{x}^{2}}{{\left( 1+x \right)}^{998}}+......+1001{{x}^{1000}}$?
Answer
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Hint: We start solving the problem by assigning a variable to the given sum. We then multiply this variable with ‘x’ and divide it with $\left( 1+x \right)$. We then subtract the obtained result from the given sum in the problem. We can then see that the obtained result has a sum resembling a sum of terms of geometric progression. We then use the fact that the sum of the terms in Geometric progression a, $ar$, $a{{r}^{2}}$,……,$a{{r}^{n-1}}$ (n terms) as $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\left( \left| r \right|<1 \right)$. We then check expansions in the obtained result in which we can get the term ${{x}^{50}}$. We then use the fact that the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1+x \right)}^{n}}\left( n\ge r \right)$ is ${}^{n}{{C}_{r}}$ to get the required result.
Complete step-by-step answer:
According to the problem, we are asked to find the coefficient of ${{x}^{50}}$ in the expression ${{\left( 1+x \right)}^{1000}}+2x{{\left( 1+x \right)}^{999}}+3{{x}^{2}}{{\left( 1+x \right)}^{998}}+......+1001{{x}^{1000}}$.
Let us assume $S={{\left( 1+x \right)}^{1000}}+2x{{\left( 1+x \right)}^{999}}+3{{x}^{2}}{{\left( 1+x \right)}^{998}}+...+1000{{x}^{999}}\left( 1+x \right)+1001{{x}^{1000}}$ ---(1).
Let us multiply both sides of equation (1) with ‘x’.
$xS=x{{\left( 1+x \right)}^{1000}}+2{{x}^{2}}{{\left( 1+x \right)}^{999}}+3{{x}^{3}}{{\left( 1+x \right)}^{998}}+...+1000{{x}^{1000}}\left( 1+x \right)+1001{{x}^{1001}}$ ---(2).
Let us divide both sides of equation (2) with $\left( 1+x \right)$.
\[\dfrac{xS}{1+x}=x{{\left( 1+x \right)}^{999}}+2{{x}^{2}}{{\left( 1+x \right)}^{998}}+3{{x}^{3}}{{\left( 1+x \right)}^{997}}+...+1000{{x}^{1000}}+\dfrac{1001{{x}^{1001}}}{1+x}\] ---(3).
Let us subtract equation (3) from equation (1).
So, we get $\Rightarrow S-\dfrac{xS}{1+x}={{\left( 1+x \right)}^{1000}}+x{{\left( 1+x \right)}^{999}}+{{x}^{2}}{{\left( 1+x \right)}^{998}}+{{x}^{3}}{{\left( 1+x \right)}^{997}}+......+{{x}^{1000}}-\dfrac{1001{{x}^{1001}}}{1+x}$.
$\Rightarrow \left( \dfrac{1+x-x}{1+x} \right)S=\dfrac{{{\left( 1+x \right)}^{1001}}+x{{\left( 1+x \right)}^{1000}}+{{x}^{2}}{{\left( 1+x \right)}^{999}}+{{x}^{3}}{{\left( 1+x \right)}^{998}}+......+{{x}^{1000}}\left( 1+x \right)-1001{{x}^{1001}}}{1+x}$.
$\Rightarrow \dfrac{S}{1+x}=\dfrac{{{\left( 1+x \right)}^{1001}}+x{{\left( 1+x \right)}^{1000}}+{{x}^{2}}{{\left( 1+x \right)}^{999}}+{{x}^{3}}{{\left( 1+x \right)}^{998}}+......+{{x}^{1000}}\left( 1+x \right)-1001{{x}^{1001}}}{1+x}$.
$\Rightarrow S={{\left( 1+x \right)}^{1001}}+x{{\left( 1+x \right)}^{1000}}+{{x}^{2}}{{\left( 1+x \right)}^{999}}+{{x}^{3}}{{\left( 1+x \right)}^{998}}+......+{{x}^{1000}}\left( 1+x \right)-1001{{x}^{1001}}$ ---(4).
Let us assume ${{S}_{1}}={{\left( 1+x \right)}^{1001}}+x{{\left( 1+x \right)}^{1000}}+{{x}^{2}}{{\left( 1+x \right)}^{999}}+{{x}^{3}}{{\left( 1+x \right)}^{998}}+......+{{x}^{1000}}\left( 1+x \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1001}}\left( 1+\dfrac{x}{1+x}+{{\left( \dfrac{x}{1+x} \right)}^{2}}+{{\left( \dfrac{x}{1+x} \right)}^{3}}+......+{{\left( \dfrac{x}{1+x} \right)}^{1000}} \right)$. We can see that this sum resembles Geometric progression with first term 1, and the common ratio $\dfrac{x}{1+x}$.
We know that the sum of the terms in Geometric progression a, $ar$, $a{{r}^{2}}$,……,$a{{r}^{n-1}}$ (n terms) is defined as $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\left( \left| r \right|<1 \right)$.
So, we get ${{S}_{1}}={{\left( 1+x \right)}^{1001}}\left( \dfrac{1\left( 1-{{\left( \dfrac{x}{1+x} \right)}^{1001}} \right)}{1-\left( \dfrac{x}{1+x} \right)} \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1001}}\left( \dfrac{1-\dfrac{{{x}^{1001}}}{{{\left( 1+x \right)}^{1001}}}}{\left( \dfrac{1+x-x}{1+x} \right)} \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1001}}\left( \dfrac{\dfrac{{{\left( 1+x \right)}^{1001}}-{{x}^{1001}}}{{{\left( 1+x \right)}^{1001}}}}{\left( \dfrac{1}{1+x} \right)} \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1001}}\left( \dfrac{{{\left( 1+x \right)}^{1001}}-{{x}^{1001}}}{{{\left( 1+x \right)}^{1000}}} \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1}}\left( {{\left( 1+x \right)}^{1001}}-{{x}^{1001}} \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1002}}-{{x}^{1001}}\left( 1+x \right)$ ---(5).
Let us substitute equation (5) in equation (1).
So, we get $S={{\left( 1+x \right)}^{1002}}-{{x}^{1001}}\left( 1+x \right)-1001{{x}^{1001}}$.
$\Rightarrow S={{\left( 1+x \right)}^{1002}}-{{x}^{1001}}-{{x}^{1002}}-1001{{x}^{1001}}$.
$\Rightarrow S={{\left( 1+x \right)}^{1002}}-1002{{x}^{1001}}-{{x}^{1002}}$ ---(6).
From equation (6), we can see that only the expansion ${{\left( 1+x \right)}^{1002}}$ has the term ${{x}^{50}}$.
We know that the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1+x \right)}^{n}}\left( n\ge r \right)$ is ${}^{n}{{C}_{r}}$.
So, the co-efficient of the term ${{x}^{50}}$ in the expansion ${{\left( 1+x \right)}^{1002}}$ is ${}^{1002}{{C}_{50}}$.
∴ The coefficient of ${{x}^{50}}$ in the expression ${{\left( 1+x \right)}^{1000}}+2x{{\left( 1+x \right)}^{999}}+3{{x}^{2}}{{\left( 1+x \right)}^{998}}+......+1001{{x}^{1000}}$ is ${}^{1002}{{C}_{50}}$.
Note: We can see that the given problems contain a huge amount of calculation so, we need to perform each step carefully in order to avoid confusion and calculation mistakes. We can see that the given expression ${{\left( 1+x \right)}^{1000}}+2x{{\left( 1+x \right)}^{999}}+3{{x}^{2}}{{\left( 1+x \right)}^{998}}+......+1001{{x}^{1000}}$ resembles the arithmetic geometric progression with common difference 1 and common ratio $\dfrac{x}{1+x}$. So, we can use the sum of ‘n’ terms of arithmetic geometric progression is $\dfrac{a}{1-r}+\dfrac{dr\left( 1-{{r}^{n-1}} \right)}{{{\left( 1-r \right)}^{2}}}-\dfrac{\left[ a+\left( n-1 \right)d \right]{{r}^{n}}}{1-r}$ to find the sum which may involve heavy calculations. Here we have assumed that $x>0$ to apply the sum of ‘n’ terms in Geometric series.
Complete step-by-step answer:
According to the problem, we are asked to find the coefficient of ${{x}^{50}}$ in the expression ${{\left( 1+x \right)}^{1000}}+2x{{\left( 1+x \right)}^{999}}+3{{x}^{2}}{{\left( 1+x \right)}^{998}}+......+1001{{x}^{1000}}$.
Let us assume $S={{\left( 1+x \right)}^{1000}}+2x{{\left( 1+x \right)}^{999}}+3{{x}^{2}}{{\left( 1+x \right)}^{998}}+...+1000{{x}^{999}}\left( 1+x \right)+1001{{x}^{1000}}$ ---(1).
Let us multiply both sides of equation (1) with ‘x’.
$xS=x{{\left( 1+x \right)}^{1000}}+2{{x}^{2}}{{\left( 1+x \right)}^{999}}+3{{x}^{3}}{{\left( 1+x \right)}^{998}}+...+1000{{x}^{1000}}\left( 1+x \right)+1001{{x}^{1001}}$ ---(2).
Let us divide both sides of equation (2) with $\left( 1+x \right)$.
\[\dfrac{xS}{1+x}=x{{\left( 1+x \right)}^{999}}+2{{x}^{2}}{{\left( 1+x \right)}^{998}}+3{{x}^{3}}{{\left( 1+x \right)}^{997}}+...+1000{{x}^{1000}}+\dfrac{1001{{x}^{1001}}}{1+x}\] ---(3).
Let us subtract equation (3) from equation (1).
So, we get $\Rightarrow S-\dfrac{xS}{1+x}={{\left( 1+x \right)}^{1000}}+x{{\left( 1+x \right)}^{999}}+{{x}^{2}}{{\left( 1+x \right)}^{998}}+{{x}^{3}}{{\left( 1+x \right)}^{997}}+......+{{x}^{1000}}-\dfrac{1001{{x}^{1001}}}{1+x}$.
$\Rightarrow \left( \dfrac{1+x-x}{1+x} \right)S=\dfrac{{{\left( 1+x \right)}^{1001}}+x{{\left( 1+x \right)}^{1000}}+{{x}^{2}}{{\left( 1+x \right)}^{999}}+{{x}^{3}}{{\left( 1+x \right)}^{998}}+......+{{x}^{1000}}\left( 1+x \right)-1001{{x}^{1001}}}{1+x}$.
$\Rightarrow \dfrac{S}{1+x}=\dfrac{{{\left( 1+x \right)}^{1001}}+x{{\left( 1+x \right)}^{1000}}+{{x}^{2}}{{\left( 1+x \right)}^{999}}+{{x}^{3}}{{\left( 1+x \right)}^{998}}+......+{{x}^{1000}}\left( 1+x \right)-1001{{x}^{1001}}}{1+x}$.
$\Rightarrow S={{\left( 1+x \right)}^{1001}}+x{{\left( 1+x \right)}^{1000}}+{{x}^{2}}{{\left( 1+x \right)}^{999}}+{{x}^{3}}{{\left( 1+x \right)}^{998}}+......+{{x}^{1000}}\left( 1+x \right)-1001{{x}^{1001}}$ ---(4).
Let us assume ${{S}_{1}}={{\left( 1+x \right)}^{1001}}+x{{\left( 1+x \right)}^{1000}}+{{x}^{2}}{{\left( 1+x \right)}^{999}}+{{x}^{3}}{{\left( 1+x \right)}^{998}}+......+{{x}^{1000}}\left( 1+x \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1001}}\left( 1+\dfrac{x}{1+x}+{{\left( \dfrac{x}{1+x} \right)}^{2}}+{{\left( \dfrac{x}{1+x} \right)}^{3}}+......+{{\left( \dfrac{x}{1+x} \right)}^{1000}} \right)$. We can see that this sum resembles Geometric progression with first term 1, and the common ratio $\dfrac{x}{1+x}$.
We know that the sum of the terms in Geometric progression a, $ar$, $a{{r}^{2}}$,……,$a{{r}^{n-1}}$ (n terms) is defined as $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\left( \left| r \right|<1 \right)$.
So, we get ${{S}_{1}}={{\left( 1+x \right)}^{1001}}\left( \dfrac{1\left( 1-{{\left( \dfrac{x}{1+x} \right)}^{1001}} \right)}{1-\left( \dfrac{x}{1+x} \right)} \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1001}}\left( \dfrac{1-\dfrac{{{x}^{1001}}}{{{\left( 1+x \right)}^{1001}}}}{\left( \dfrac{1+x-x}{1+x} \right)} \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1001}}\left( \dfrac{\dfrac{{{\left( 1+x \right)}^{1001}}-{{x}^{1001}}}{{{\left( 1+x \right)}^{1001}}}}{\left( \dfrac{1}{1+x} \right)} \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1001}}\left( \dfrac{{{\left( 1+x \right)}^{1001}}-{{x}^{1001}}}{{{\left( 1+x \right)}^{1000}}} \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1}}\left( {{\left( 1+x \right)}^{1001}}-{{x}^{1001}} \right)$.
$\Rightarrow {{S}_{1}}={{\left( 1+x \right)}^{1002}}-{{x}^{1001}}\left( 1+x \right)$ ---(5).
Let us substitute equation (5) in equation (1).
So, we get $S={{\left( 1+x \right)}^{1002}}-{{x}^{1001}}\left( 1+x \right)-1001{{x}^{1001}}$.
$\Rightarrow S={{\left( 1+x \right)}^{1002}}-{{x}^{1001}}-{{x}^{1002}}-1001{{x}^{1001}}$.
$\Rightarrow S={{\left( 1+x \right)}^{1002}}-1002{{x}^{1001}}-{{x}^{1002}}$ ---(6).
From equation (6), we can see that only the expansion ${{\left( 1+x \right)}^{1002}}$ has the term ${{x}^{50}}$.
We know that the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1+x \right)}^{n}}\left( n\ge r \right)$ is ${}^{n}{{C}_{r}}$.
So, the co-efficient of the term ${{x}^{50}}$ in the expansion ${{\left( 1+x \right)}^{1002}}$ is ${}^{1002}{{C}_{50}}$.
∴ The coefficient of ${{x}^{50}}$ in the expression ${{\left( 1+x \right)}^{1000}}+2x{{\left( 1+x \right)}^{999}}+3{{x}^{2}}{{\left( 1+x \right)}^{998}}+......+1001{{x}^{1000}}$ is ${}^{1002}{{C}_{50}}$.
Note: We can see that the given problems contain a huge amount of calculation so, we need to perform each step carefully in order to avoid confusion and calculation mistakes. We can see that the given expression ${{\left( 1+x \right)}^{1000}}+2x{{\left( 1+x \right)}^{999}}+3{{x}^{2}}{{\left( 1+x \right)}^{998}}+......+1001{{x}^{1000}}$ resembles the arithmetic geometric progression with common difference 1 and common ratio $\dfrac{x}{1+x}$. So, we can use the sum of ‘n’ terms of arithmetic geometric progression is $\dfrac{a}{1-r}+\dfrac{dr\left( 1-{{r}^{n-1}} \right)}{{{\left( 1-r \right)}^{2}}}-\dfrac{\left[ a+\left( n-1 \right)d \right]{{r}^{n}}}{1-r}$ to find the sum which may involve heavy calculations. Here we have assumed that $x>0$ to apply the sum of ‘n’ terms in Geometric series.
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