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

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:

& \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{aligned}$$ So here we see that we have $$\sum\limits_{r=0}^{n}{{{(-1)}^{n}}{}^{n}{{C}_{r}}}={{\left( -2 \right)}^{n}}$$,so above expression is: $$\begin{aligned} & \Rightarrow \dfrac{1}{n!}\sum\limits_{r=0}^{n}{{{(-1)}^{n}}{}^{n}{{C}_{r}}} \\\ & \Rightarrow \dfrac{1}{n!}{{\left( -2 \right)}^{n}} \\\ \end{aligned}$$ So finally, we can say that the value of the expression $$\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.