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Question: If \( n \geqslant 2\) then \(3{C_1} - 4{C_2} + 5{C_3} - .......1{\left( { - 1} \right)^{n - 1}}\left...

If n⩾2 n \geqslant 2 then 3C1−4C2+5C3−.......1(−1)n−1(n+2)Cn3{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 - 1
B. 2
C.−2 - 2
D.1

Explanation

Solution

We got the nth{n^{th}} term as (−1)n−1(n+2)Cn{\left( { - 1} \right)^{n - 1}}\left( {n + 2} \right){C_n} and the given series is
i=n∑i=2(−1)n−1(i+2)nCi\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=−1x = - 1 for (1+x)n{\left( {1 + x} \right)^n}

Complete step-by-step answer:
Let’s begin with given expansion. It is
⇒3nC1−4nC2+5nC3−........+(−1)n−1(n+2)nCn\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 nth{n^{th}} term as
⇒(−1)r−1(r+2)  nCr\Rightarrow {\left( { - 1} \right)^{r - 1}}\left( {r + 2} \right){\;^n}{C_r}
Hence, series can also be written as
n∑r=1(−1)n−1(r+2)  nCr\dfrac{n}{{\dfrac{\sum }{{r = 1}}}}{\left( { - 1} \right)^{n - 1}}\left( {r + 2} \right){\;^n}{C_r} n∑r=1[(−1)r−1r  nCr+(−1)r−12.  2  nCr]\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 n∑r=1(−1)r−1r.  nCr      & 2.n∑r=1  nCr\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 n∑i=1(−1)r  r  ncr\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
(1+x)n=nC0+nC1x+nC2x2+nC3x3+......+nCnxn{\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  xw.r.t\;x
⇒n(1+x)n−1=0+nC1+nC2.2x+nC3.3x2+.....nCn∗n.xn−1\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}}
⇒n(1+x)n−1=nC1+2n.C2x+3.nC3.x2+......−1n.nCn.xn−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 (−1)\left( { - 1} \right) in each term we can use x=−1.x = - 1. so we get.
⇒n(1−1)n−1=0+nC1+2.nC2(−1)+3.nC3(−1)2+......+n.nCn(−1)\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)
⇒0=0+nC1−2.Cn2+3.nC3−4.nC4......\Rightarrow 0 = 0{ + ^n}{C_1} - {2.^C}{n_2} + {3.^n}{C_3} - {4.^n}{C_4}......
Hence. We got
n∑r=1(−1)r=1r.nCr=0\dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^{r = 1}}r{.^n}{C_r} = 0 ①
Now, let’s compute
n∑r=1(−1)xCr,\dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^x}{C_r},
Which can be computed from Coefficient of (1+x)n{\left( {1 + x} \right)^ n } Binomial expansion will be
(1+x)n=nC0+nC1x+nC2x2+......+nCnxn{\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
[nC1−nC2+nC3−nC4+.........]\left[ {^n{C_1}{ - ^n}{C_2}{ + ^n}{C_3}{ - ^n}{C_4} + .........} \right]
We need to put a Value of x=−1x = - 1 . Therefore the equation will give
⇒(1−1)n=nC0+nC1(−1)+nC2(−1)2+........+nCn(−1)n\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}
⇒0=nC0−nC1+nC2−nC3+.........+nCn(−1)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
⇒0=nC0−[nC1−nC2+nC3−nC4+......+(−1)n−1nCn]\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]
=nC0−n∑r=1(−1)r−1nCr{ = ^n}{C_0} - \dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^{r - 1}}^n{C_r}
Therefore, n∑r=1(−1)n−1nCr=nC0=1\dfrac{{\dfrac{n}{\sum }}}{{r = 1}}{\left( { - 1} \right)^{n - 1}}^n{C_r}{ = ^n}{C_0} = 1
We required 2×n∑r=1(−1)n−1nCr=2×(1)=22 \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
⇒n∑r=1(−1)r−1(r+1)nCr+2n∑r=1(−1)r−1nCr\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
⇒0+2(1)\Rightarrow 0 + 2\left( 1 \right)
=2= 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.
⇒nCr+nCr−1=n+1Cr{ \Rightarrow ^n}{C_r}{ + ^n}{C_{r - 1}}{ = ^{n + 1}}{C_r}