Find the coefficient of x in the expansion of \[\left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}\]
Answer
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Hint: First we should expand the expansion \[{{\left( 1-x \right)}^{16}}\]. We know that \[{{\left( 1-x \right)}^{n}}=1{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}{{-}^{n}}{{C}_{3}}{{x}^{3}}+......+{{\left( -1 \right)}^{r}}^{n}{{C}_{r}}{{x}^{r}}+.....+{{\left( -1 \right)}^{n}}^{n}{{C}_{n}}{{x}^{n}}\]. By using this formula, we should expand the expansion \[{{\left( 1-x \right)}^{16}}\]. Now we should expand the expansion \[\left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}\]. Now we should separate the coefficients of \[\left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}\] and this should be written in the form of \[1+ax+b{{x}^{2}}+c{{x}^{3}}+.......\]. Now we should write the coefficient of x. We know that \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]. Now, we should use this formula, to find the coefficient of x.
Complete step-by-step answer:
Before solving the question, we should know that \[{{\left( 1-x \right)}^{n}}=1{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}{{-}^{n}}{{C}_{3}}{{x}^{3}}+......+{{\left( -1 \right)}^{r}}^{n}{{C}_{r}}{{x}^{r}}+.....+{{\left( -1 \right)}^{n}}^{n}{{C}_{n}}{{x}^{n}}\].
From the question, it is given that we should find the coefficient of x in the expansion of \[\left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}\].
We know that \[{{\left( 1-x \right)}^{n}}=1{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}{{-}^{n}}{{C}_{3}}{{x}^{3}}+......+{{\left( -1 \right)}^{r}}^{n}{{C}_{r}}{{x}^{r}}+.....+{{\left( -1 \right)}^{n}}^{n}{{C}_{n}}{{x}^{n}}\]
\[\Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( 1-3x+7{{x}^{2}} \right)\left( 1{{-}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{-}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}-...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right)\]
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}}{{-}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{-}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}-...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ }-3x\left( ^{16}{{C}_{0}}{{-}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{-}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}-...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ +7}{{\text{x}}^{2}}\left( ^{16}{{C}_{0}}{{-}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{-}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}-...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}}{{-}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{-}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}-...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ }-3\left( ^{16}{{C}_{0}}x{{-}^{16}}{{C}_{1}}{{x}^{2}}{{+}^{16}}{{C}_{2}}{{x}^{3}}{{-}^{16}}{{C}_{3}}{{x}^{4}}{{+}^{16}}{{C}_{4}}{{x}^{5}}-...........{{+}^{16}}{{C}_{16}}{{x}^{17}} \right) \\
& \text{ +7}\left( ^{16}{{C}_{0}}{{x}^{2}}{{-}^{16}}{{C}_{1}}{{x}^{3}}{{+}^{16}}{{C}_{2}}{{x}^{4}}{{-}^{16}}{{C}_{3}}{{x}^{5}}{{+}^{16}}{{C}_{4}}{{x}^{6}}-...........{{+}^{16}}{{C}_{16}}{{x}^{18}} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}} \right)-\left( ^{16}{{C}_{1}}+{{3}^{16}}{{C}_{0}} \right)x+\left( ^{16}{{C}_{2}}+{{3}^{16}}{{C}_{1}}+{{7}^{16}}{{C}_{0}} \right){{x}^{2}}-\left( ^{16}{{C}_{3}}+{{3}^{16}}{{C}_{2}}+{{7}^{16}}{{C}_{1}} \right){{x}^{3}} \\
& \text{ }+\left( ^{16}{{C}_{4}}+{{3}^{16}}{{C}_{4}}+{{7}^{16}}{{C}_{2}} \right){{x}^{4}}-......-\left( ^{16}{{C}_{16}}+{{3}^{16}}{{C}_{15}}+{{7}^{16}}{{C}_{14}} \right){{x}^{16}} \\
& \text{ +}\left( {{3}^{16}}{{C}_{16}}+{{7}^{16}}{{C}_{15}} \right){{x}^{17}}-\left( {{7}^{16}}{{C}_{15}} \right){{x}^{18}} \\
\end{align}\]From the question, it is clear that we should find the coefficient of x.
So, let us assume the coefficient of x is equal to T.
\[\Rightarrow T=\left( -1 \right)\left( ^{16}{{C}_{1}}+{{3}^{16}}{{C}_{0}} \right)\]
We know that \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
\[\begin{align}
& \Rightarrow T=\left( -1 \right)\left( \dfrac{16!}{0!\left( 16-1 \right)!}+3\left( \dfrac{16!}{0!\left( 16-0 \right)!} \right) \right) \\
& \Rightarrow T=\left( -1 \right)\left( \dfrac{16!}{\left( 15 \right)!}+3\left( \dfrac{16!}{\left( 16 \right)!} \right) \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow T=\left( -1 \right)\left( \dfrac{16.15!}{\left( 15 \right)!}+3\left( \dfrac{16!}{\left( 16 \right)!} \right) \right) \\
& \Rightarrow T=-1\left( 16+3\left( 1 \right) \right) \\
& \Rightarrow T=-1\left( 16+3 \right) \\
& \Rightarrow T=-19......(1) \\
\end{align}\]
So, from equation (1) it is clear that the coefficient of x in the expansion of \[\left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}\] is equal to -19.
Note: Students may have a misconception that \[{{\left( 1-x \right)}^{n}}=1{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}{{+}^{n}}{{C}_{3}}{{x}^{3}}+......{{+}^{n}}{{C}_{r}}{{x}^{r}}+.....{{+}^{n}}{{C}_{n}}{{x}^{n}}\]. If this misconception is followed, then we will get a wrong answer as shown below:
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}}{{+}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{+}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}+...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ }-3x\left( ^{16}{{C}_{0}}{{+}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{+}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}+...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ +7}{{\text{x}}^{2}}\left( ^{16}{{C}_{0}}{{+}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{+}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}+...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}}{{+}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{+}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}+...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ }-3\left( ^{16}{{C}_{0}}x{{+}^{16}}{{C}_{1}}{{x}^{2}}{{+}^{16}}{{C}_{2}}{{x}^{3}}{{+}^{16}}{{C}_{3}}{{x}^{4}}{{+}^{16}}{{C}_{4}}{{x}^{5}}+...........{{+}^{16}}{{C}_{16}}{{x}^{17}} \right) \\
& \text{ +7}\left( ^{16}{{C}_{0}}{{x}^{2}}{{+}^{16}}{{C}_{1}}{{x}^{3}}{{+}^{16}}{{C}_{2}}{{x}^{4}}{{+}^{16}}{{C}_{3}}{{x}^{5}}{{+}^{16}}{{C}_{4}}{{x}^{6}}+...........{{+}^{16}}{{C}_{16}}{{x}^{18}} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}} \right)+\left( ^{16}{{C}_{1}}-{{3}^{16}}{{C}_{0}} \right)x+\left( ^{16}{{C}_{2}}-{{3}^{16}}{{C}_{1}}+{{7}^{16}}{{C}_{0}} \right){{x}^{2}}+\left( ^{16}{{C}_{3}}-{{3}^{16}}{{C}_{2}}+{{7}^{16}}{{C}_{1}} \right){{x}^{3}} \\
& \text{ }+\left( ^{16}{{C}_{4}}-{{3}^{16}}{{C}_{4}}+{{7}^{16}}{{C}_{2}} \right){{x}^{4}}+......+\left( ^{16}{{C}_{16}}-{{3}^{16}}{{C}_{15}}+{{7}^{16}}{{C}_{14}} \right){{x}^{16}} \\
& \text{ +}\left( -{{3}^{16}}{{C}_{16}}+{{7}^{16}}{{C}_{15}} \right){{x}^{17}}+\left( {{7}^{16}}{{C}_{15}} \right){{x}^{18}} \\
\end{align}\]
From the question, it is clear that we should find the coefficient of x.
So, let us assume the coefficient of x is equal to T.
\[\Rightarrow T=\left( ^{16}{{C}_{1}}-{{3}^{16}}{{C}_{0}} \right)\]
We know that \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
\[\begin{align}
& \Rightarrow T=\left( \dfrac{16!}{0!\left( 16-1 \right)!}-3\left( \dfrac{16!}{0!\left( 16-0 \right)!} \right) \right) \\
& \Rightarrow T=\left( \dfrac{16!}{\left( 15 \right)!}-3\left( \dfrac{16!}{\left( 16 \right)!} \right) \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow T=\left( \dfrac{16.15!}{\left( 15 \right)!}-3\left( \dfrac{16!}{\left( 16 \right)!} \right) \right) \\
& \Rightarrow T=\left( 16-3\left( 1 \right) \right) \\
& \Rightarrow T=\left( 16-3 \right) \\
& \Rightarrow T=13......(1) \\
\end{align}\]
So, from equation (1) it is clear that the coefficient of x in the expansion of \[\left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}\] is equal to 13.
But we know that the coefficient of x is equal to -19 but we got the coefficient of x is equal to 13. So, this misconception should be avoided.
Complete step-by-step answer:
Before solving the question, we should know that \[{{\left( 1-x \right)}^{n}}=1{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}{{-}^{n}}{{C}_{3}}{{x}^{3}}+......+{{\left( -1 \right)}^{r}}^{n}{{C}_{r}}{{x}^{r}}+.....+{{\left( -1 \right)}^{n}}^{n}{{C}_{n}}{{x}^{n}}\].
From the question, it is given that we should find the coefficient of x in the expansion of \[\left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}\].
We know that \[{{\left( 1-x \right)}^{n}}=1{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}{{-}^{n}}{{C}_{3}}{{x}^{3}}+......+{{\left( -1 \right)}^{r}}^{n}{{C}_{r}}{{x}^{r}}+.....+{{\left( -1 \right)}^{n}}^{n}{{C}_{n}}{{x}^{n}}\]
\[\Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( 1-3x+7{{x}^{2}} \right)\left( 1{{-}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{-}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}-...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right)\]
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}}{{-}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{-}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}-...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ }-3x\left( ^{16}{{C}_{0}}{{-}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{-}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}-...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ +7}{{\text{x}}^{2}}\left( ^{16}{{C}_{0}}{{-}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{-}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}-...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}}{{-}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{-}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}-...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ }-3\left( ^{16}{{C}_{0}}x{{-}^{16}}{{C}_{1}}{{x}^{2}}{{+}^{16}}{{C}_{2}}{{x}^{3}}{{-}^{16}}{{C}_{3}}{{x}^{4}}{{+}^{16}}{{C}_{4}}{{x}^{5}}-...........{{+}^{16}}{{C}_{16}}{{x}^{17}} \right) \\
& \text{ +7}\left( ^{16}{{C}_{0}}{{x}^{2}}{{-}^{16}}{{C}_{1}}{{x}^{3}}{{+}^{16}}{{C}_{2}}{{x}^{4}}{{-}^{16}}{{C}_{3}}{{x}^{5}}{{+}^{16}}{{C}_{4}}{{x}^{6}}-...........{{+}^{16}}{{C}_{16}}{{x}^{18}} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}} \right)-\left( ^{16}{{C}_{1}}+{{3}^{16}}{{C}_{0}} \right)x+\left( ^{16}{{C}_{2}}+{{3}^{16}}{{C}_{1}}+{{7}^{16}}{{C}_{0}} \right){{x}^{2}}-\left( ^{16}{{C}_{3}}+{{3}^{16}}{{C}_{2}}+{{7}^{16}}{{C}_{1}} \right){{x}^{3}} \\
& \text{ }+\left( ^{16}{{C}_{4}}+{{3}^{16}}{{C}_{4}}+{{7}^{16}}{{C}_{2}} \right){{x}^{4}}-......-\left( ^{16}{{C}_{16}}+{{3}^{16}}{{C}_{15}}+{{7}^{16}}{{C}_{14}} \right){{x}^{16}} \\
& \text{ +}\left( {{3}^{16}}{{C}_{16}}+{{7}^{16}}{{C}_{15}} \right){{x}^{17}}-\left( {{7}^{16}}{{C}_{15}} \right){{x}^{18}} \\
\end{align}\]From the question, it is clear that we should find the coefficient of x.
So, let us assume the coefficient of x is equal to T.
\[\Rightarrow T=\left( -1 \right)\left( ^{16}{{C}_{1}}+{{3}^{16}}{{C}_{0}} \right)\]
We know that \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
\[\begin{align}
& \Rightarrow T=\left( -1 \right)\left( \dfrac{16!}{0!\left( 16-1 \right)!}+3\left( \dfrac{16!}{0!\left( 16-0 \right)!} \right) \right) \\
& \Rightarrow T=\left( -1 \right)\left( \dfrac{16!}{\left( 15 \right)!}+3\left( \dfrac{16!}{\left( 16 \right)!} \right) \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow T=\left( -1 \right)\left( \dfrac{16.15!}{\left( 15 \right)!}+3\left( \dfrac{16!}{\left( 16 \right)!} \right) \right) \\
& \Rightarrow T=-1\left( 16+3\left( 1 \right) \right) \\
& \Rightarrow T=-1\left( 16+3 \right) \\
& \Rightarrow T=-19......(1) \\
\end{align}\]
So, from equation (1) it is clear that the coefficient of x in the expansion of \[\left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}\] is equal to -19.
Note: Students may have a misconception that \[{{\left( 1-x \right)}^{n}}=1{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}{{+}^{n}}{{C}_{3}}{{x}^{3}}+......{{+}^{n}}{{C}_{r}}{{x}^{r}}+.....{{+}^{n}}{{C}_{n}}{{x}^{n}}\]. If this misconception is followed, then we will get a wrong answer as shown below:
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}}{{+}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{+}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}+...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ }-3x\left( ^{16}{{C}_{0}}{{+}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{+}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}+...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ +7}{{\text{x}}^{2}}\left( ^{16}{{C}_{0}}{{+}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{+}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}+...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}}{{+}^{16}}{{C}_{1}}x{{+}^{16}}{{C}_{2}}{{x}^{2}}{{+}^{16}}{{C}_{3}}{{x}^{3}}{{+}^{16}}{{C}_{4}}{{x}^{4}}+...........{{+}^{16}}{{C}_{16}}{{x}^{16}} \right) \\
& \text{ }-3\left( ^{16}{{C}_{0}}x{{+}^{16}}{{C}_{1}}{{x}^{2}}{{+}^{16}}{{C}_{2}}{{x}^{3}}{{+}^{16}}{{C}_{3}}{{x}^{4}}{{+}^{16}}{{C}_{4}}{{x}^{5}}+...........{{+}^{16}}{{C}_{16}}{{x}^{17}} \right) \\
& \text{ +7}\left( ^{16}{{C}_{0}}{{x}^{2}}{{+}^{16}}{{C}_{1}}{{x}^{3}}{{+}^{16}}{{C}_{2}}{{x}^{4}}{{+}^{16}}{{C}_{3}}{{x}^{5}}{{+}^{16}}{{C}_{4}}{{x}^{6}}+...........{{+}^{16}}{{C}_{16}}{{x}^{18}} \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow \left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}=\left( ^{16}{{C}_{0}} \right)+\left( ^{16}{{C}_{1}}-{{3}^{16}}{{C}_{0}} \right)x+\left( ^{16}{{C}_{2}}-{{3}^{16}}{{C}_{1}}+{{7}^{16}}{{C}_{0}} \right){{x}^{2}}+\left( ^{16}{{C}_{3}}-{{3}^{16}}{{C}_{2}}+{{7}^{16}}{{C}_{1}} \right){{x}^{3}} \\
& \text{ }+\left( ^{16}{{C}_{4}}-{{3}^{16}}{{C}_{4}}+{{7}^{16}}{{C}_{2}} \right){{x}^{4}}+......+\left( ^{16}{{C}_{16}}-{{3}^{16}}{{C}_{15}}+{{7}^{16}}{{C}_{14}} \right){{x}^{16}} \\
& \text{ +}\left( -{{3}^{16}}{{C}_{16}}+{{7}^{16}}{{C}_{15}} \right){{x}^{17}}+\left( {{7}^{16}}{{C}_{15}} \right){{x}^{18}} \\
\end{align}\]
From the question, it is clear that we should find the coefficient of x.
So, let us assume the coefficient of x is equal to T.
\[\Rightarrow T=\left( ^{16}{{C}_{1}}-{{3}^{16}}{{C}_{0}} \right)\]
We know that \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
\[\begin{align}
& \Rightarrow T=\left( \dfrac{16!}{0!\left( 16-1 \right)!}-3\left( \dfrac{16!}{0!\left( 16-0 \right)!} \right) \right) \\
& \Rightarrow T=\left( \dfrac{16!}{\left( 15 \right)!}-3\left( \dfrac{16!}{\left( 16 \right)!} \right) \right) \\
\end{align}\]
\[\begin{align}
& \Rightarrow T=\left( \dfrac{16.15!}{\left( 15 \right)!}-3\left( \dfrac{16!}{\left( 16 \right)!} \right) \right) \\
& \Rightarrow T=\left( 16-3\left( 1 \right) \right) \\
& \Rightarrow T=\left( 16-3 \right) \\
& \Rightarrow T=13......(1) \\
\end{align}\]
So, from equation (1) it is clear that the coefficient of x in the expansion of \[\left( 1-3x+7{{x}^{2}} \right){{\left( 1-x \right)}^{16}}\] is equal to 13.
But we know that the coefficient of x is equal to -19 but we got the coefficient of x is equal to 13. So, this misconception should be avoided.
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