Question

Find the coefficient of \[{{x}^{n}}\]in \[{{\left( 1-x+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+.....+\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!} \right)}^{2}}\]. Choose the correct option.
A. \[\dfrac{{{\left( -n \right)}^{n}}}{n!}\]
B. \[\dfrac{{{\left( -2 \right)}^{n}}}{n!}\]
C. \[\dfrac{1}{{{\left( n! \right)}^{2}}}\]
D. \[-\dfrac{1}{{{\left( n! \right)}^{2}}}\]

Answer 1

Hint: First multiply \[\left( 1-x+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+.....+\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!} \right)\times \left( 1-x+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+.....+\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!} \right)\]. Now, here \[{{x}^{n}}\] is found by multiplying \[1\times {{x}^{n}}={{x}^{n}}\]or \[x\times {{x}^{n-1}}={{x}^{n}}\]or \[{{x}^{2}}\times {{x}^{n-2}}={{x}^{n}}\]and so on. So add all the coefficients of \[{{x}^{n}}\]and then use the formula that \[{{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{a}^{\left( n-r \right)}}{{b}^{r}}\], when \[a=b=-1\], we have: \[{{\left( -2 \right)}^{n}}=\sum\limits_{r=0}^{n}{{{(-1)}^{n}}{}^{n}{{C}_{r}}}\]

Complete step-by-step answer:
In the question, we have to find the coefficient of \[{{x}^{n}}\]in \[{{\left( 1-x+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+.....+\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!} \right)}^{2}}\].
Now we can write\[{{\left( 1-x+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+.....+\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!} \right)}^{2}}=\left( 1-x+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+.....+\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!} \right)\times \left( 1-x+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+.....+\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!} \right)\]
Next, we can see that when we multiply \[1\times \dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!}=\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!}\]has the \[{{x}^{n}}\], similarly when we multiply \[-x\times \dfrac{{{\left( -1 \right)}^{n-1}}{{x}^{n-1}}}{(n-1)!}=\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{(n-1)!}\] will gain have the \[{{x}^{n}}\]. Also, when we multiply \[\dfrac{{{x}^{2}}}{2!}\times \dfrac{{{\left( -1 \right)}^{n}}{{x}^{n-2}}}{(n-2)!}=\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{2!(n-2)!}\]will again have \[{{x}^{n}}\].
So, we will have the pattern \[\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{0!n!}+\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{2!(n-2)!}+......\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{0!n!}\]add together will have the \[{{x}^{n}}\]. So, to find the coefficients \[{{x}^{n}}\]of we have to add all the terms \[\dfrac{{{\left( -1 \right)}^{n}}}{0!n!}+\dfrac{{{\left( -1 \right)}^{n}}}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}}{2!(n-2)!}+......\dfrac{{{\left( -1 \right)}^{n}}}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}}{0!n!}\]
Now, we know that that expansion formula for the expression of the form \[{{\left( a+b \right)}^{n}}\]will be given by \[{{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{a}^{\left( n-r \right)}}{{b}^{r}}\], So here when \[a=b=-1\], we have: \[{{\left( -2 \right)}^{n}}=\sum\limits_{r=0}^{n}{{{(-1)}^{n}}{}^{n}{{C}_{r}}}\]. Now, we have to find the value of the expression \[\dfrac{{{\left( -1 \right)}^{n}}}{0!n!}+\dfrac{{{\left( -1 \right)}^{n}}}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}}{2!(n-2)!}+......\dfrac{{{\left( -1 \right)}^{n}}}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}}{0!n!}\], so this can be done as follows:
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( -1 \right)}^{n}}}{0!n!}+\dfrac{{{\left( -1 \right)}^{n}}}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}}{2!(n-2)!}+......\dfrac{{{\left( -1 \right)}^{n}}}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}}{0!n!} \\
 & \Rightarrow \dfrac{1}{n!}\left( \dfrac{{{\left( -1 \right)}^{n}}n!}{0!n!}+\dfrac{{{\left( -1 \right)}^{n}}n!}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}n!}{2!(n-2)!}+......\dfrac{{{\left( -1 \right)}^{n}}n!}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}n!}{0!n!} \right) \\
 & \Rightarrow \dfrac{1}{n!}\sum\limits_{r=0}^{n}{{{(-1)}^{n}}{}^{n}{{C}_{r}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because {}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!} \\
\end{align}\]
So here we see that we have \[\sum\limits_{r=0}^{n}{{{(-1)}^{n}}{}^{n}{{C}_{r}}}={{\left( -2 \right)}^{n}}\],so above expression is:
\[\begin{align}
  & \Rightarrow \dfrac{1}{n!}\sum\limits_{r=0}^{n}{{{(-1)}^{n}}{}^{n}{{C}_{r}}} \\
 & \Rightarrow \dfrac{1}{n!}{{\left( -2 \right)}^{n}} \\
\end{align}\]
 So finally, we can say that the value of the expression \[\dfrac{{{\left( -1 \right)}^{n}}}{0!n!}+\dfrac{{{\left( -1 \right)}^{n}}}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}}{2!(n-2)!}+......\dfrac{{{\left( -1 \right)}^{n}}}{1!(n-1)!}+\dfrac{{{\left( -1 \right)}^{n}}}{0!n!}=\dfrac{{{\left( -2 \right)}^{n}}}{n!}\]
Hence, this is the value of the coefficient of \[{{x}^{n}}\]. So, the correct answer is option B.

Note: It can be noted that \[0!=1!=1\], so we have to be careful in that. Also we should be careful that we have \[{{\left( \sum\limits_{n=0}^{\infty }{\;}\dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!} \right)}^{2}}\ne \left( \sum\limits_{n=0}^{\infty }{\;}{{\left( \dfrac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!} \right)}^{2}} \right)\]. Make a note of the expansion formula \[{{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{a}^{\left( n-r \right)}}{{b}^{r}}\] is applicable to all types of constant or a variable, integer or a fractional number.