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If $ n \geqslant 2$ then $3{C_1} - 4{C_2} + 5{C_3} - .......1{\left( { - 1} \right)^{n - 1}}\left( {n + 2} \right){C_n}$ is equal to
A.$ - 1$
B. 2
C.$ - 2$
D.1

Answer Verified Verified
Hint: We got the ${n^{th}}$ term as \[{\left( { - 1} \right)^{n - 1}}\left( {n + 2} \right){C_n}\] and the given series is
     \[\dfrac{{i = n}}{{\dfrac{\sum }{{i = 2}}}}{\left( { - 1} \right)^{n - 1}}{\left( {i + 2} \right)^n}Ci\]
Divide the express into 2 and relate it with binomial expansion’s coefficient where $x = - 1$ for ${\left( {1 + x} \right)^n}$

Complete step-by-step answer:
Let’s begin with given expansion. It is
\[ \Rightarrow {3^n}{C_1} - {4^ n }{C_2} + {5^n}{C_3} - ........ + {\left( { - 1} \right)^{n - 1}}{\left( {n + 2} \right)^n}{C_ n}\]
So we can take ${n^{th}}$ term as
$ \Rightarrow {\left( { - 1} \right)^{r - 1}}\left( {r + 2} \right){\;^n}{C_r}$
Hence, series can also be written as
$\dfrac{n}{{\dfrac{\sum }{{r = 1}}}}{\left( { - 1} \right)^{n - 1}}\left( {r + 2} \right){\;^n}{C_r}$ $\dfrac{{\dfrac{n}{\sum }}}{{r = 1}}\left[ {{{\left( { - 1} \right)}^{r - 1}}r{\;^n}{C_r} + {{\left( { - 1} \right)}^{r - 1}}2.\;2{\;^n}{C_r}} \right]$
So we can divide the expression into Z. where $\dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^{r - 1}}r.{\;^n}{C_r}\;\;\;\& \,2.\dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\;^n}{C_r}$
First, I would like to solve $\dfrac{{\dfrac{n}{\sum }}}{{i = 1}}{\left( { - 1} \right)^r}\;r{\;^n}{c_r}$. By observing the equation formation. Each of the values having term r as a multiple is quite different from a regular expression. It can be obtained on a regular basis , if the variable is differentiator.
As we know that binomial expansion of
${\left( {1 + x} \right)^ n }{ = ^n}{C_0}{ + ^n}{C_1}x{ + ^n}{C_2}{x^2}{ + ^n}{C_3}{x^3} + ......{ + ^n}{C_n}{x^n}$
If we differentiate both the side, we get $w.r.t\;x$
\[ \Rightarrow n{\left( {1 + x} \right)^{n - 1}} = 0{ + ^n}{C_1}{ + ^n}{C_2}.2x{ + ^n}{C_3}.3{x^{^2}} + {.....^n}{C_n} * n.{x^{n - 1}}\]
\[ \Rightarrow n{\left( {1 + x} \right)^{n - 1}}{ = ^n}{C_1} + {2^n}.{C_2}x + {3.^n}{C_3}.{x^2} + ...... - 1n{.^n}{C_n}.{x^{n - 1}}\]
To obtain the relation with $\left( { - 1} \right)$ in each term we can use $x = - 1.$ so we get.
$ \Rightarrow n{\left( {1 - 1} \right)^{n - 1}} = 0{ + ^n}{C_1} + {2.^n}{C_2}\left( { - 1} \right) + {3.^n}{C_3}{\left( { - 1} \right)^2} + ...... + n{.^n}{C_n}\left( { - 1} \right)$
$ \Rightarrow 0 = 0{ + ^n}{C_1} - {2.^C}{n_2} + {3.^n}{C_3} - {4.^n}{C_4}......$
Hence. We got
$\dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^{r = 1}}r{.^n}{C_r} = 0$ ①
Now, let’s compute
\[\dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^x}{C_r},\]
Which can be computed from Coefficient of ${\left( {1 + x} \right)^ n }$ Binomial expansion will be
${\left( {1 + x} \right)^n}{ = ^n}{C_0}{ + ^n}{C_1}x{ + ^n}{C_2}{x^2} + ......{ + ^n}{C_n}{x^n}$
If if need the expression in the form
\[\left[ {^n{C_1}{ - ^n}{C_2}{ + ^n}{C_3}{ - ^n}{C_4} + .........} \right]\]
We need to put a Value of $x = - 1$ . Therefore the equation will give
$ \Rightarrow {\left( {1 - 1} \right)^n}{ = ^n}{C_0}{ + ^n}{C_1}\left( { - 1} \right){ + ^n}{C_2}{\left( { - 1} \right)^2} + ........{ + ^n}{C_n}{\left( { - 1} \right)^n}$
$ \Rightarrow 0{ = ^n}{C_0}{ - ^n}{C_1}{ + ^n}{C_2}{ - ^n}{C_3} + .........{ + ^n}{C_n}{\left( { - 1} \right)^n}$
If we compare, the equation resulted in
 $ \Rightarrow 0{ = ^n}{C_0} - \left[ {^n{C_1}{ - ^n}{C_2}{ + ^n}{C_3}{ - ^n}{C_4} + ...... + {{\left( { - 1} \right)}^{n - 1}}^n{C_n}} \right]$
     ${ = ^n}{C_0} - \dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^{r - 1}}^n{C_r}$
Therefore, $\dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^{n - 1}}^n{C_r}{ = ^n}{C_0} = 1$
We required $2 \times \dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^{n - 1}}^n{C_r} = 2 \times \left( 1 \right) = 2$ (2)
Hence we got both the value. So the equation given
$ \Rightarrow \dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^{r - 1}}{\left( {r + 1} \right)^n}{C_r} + 2\dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^{r - 1}}^n{C_r}$
Using (1) and (2) we got
$ \Rightarrow 0 + 2\left( 1 \right)$
$ = 2$

Hence, option B is the correct answer.

Note: If we get into any binomial form of expression. It will be anyhow, the form of Binomial Expression for any short Expression. Binomial Properties are used to shorten the calculation like.
${ \Rightarrow ^n}{C_r}{ + ^n}{C_{r - 1}}{ = ^{n + 1}}{C_r}$