Question

If in the expansion of ${\left(1-x\right)}^{2n-1}$, the coefficient of $x^r$ denoted by $a_r$, then :

• A. $a_{r-1}+a_{2n-r}=0$
• B. $a_{r-1}-a_{2n-r}=0$
• C. $a_{r-1}+2a_{2n-r}=0$
• D. None of these

Answer 1

The correct option is C $a_{r-1}+2a_{2n-r}=0$

We have,

$ar-1=$ coefficient of $x^{r-1}in(1-x{)}^{2n-1}=(-1{)}^{r-1},2^{n-1}Cr-1$

$a2n-r$ coefficient of $c^{2n-r}in(1-x{)}^{2n-1}=(-1{)}^{2n-r}{.2}^{n-1}C2n-r$

Now, $ar-1+a2n-r=(-1{)}^{r-1}{.}^{2n-1}Cr-1+(-1{)}^{2n-r}.2^{n-1}{2}^{n-1}C2n-r$

$=(-1{)}^{r-1}{.2}^{n-1}C(2n-1)-(r-1)+(-1{)}^{2n}.(-1{)}^{-r}{.}^{2n-1}C2n-r[as{}^n{C}_r{-}^{n}C-r]$

$(-1{)}^{r-1}{.}^{2n-1}C2n-r+(-1{)}^{-r}{.}^{2n-1}C2n-r$

${=}^{2n-1}C2n-r\left[\right(-1{)}^{r-1}+(-1{)}^{-r}]$

${\left[\right(-1{)}^{r-1}+\frac{1}{(-1{)}^r}]}^{2N-1}C2n-r$

${\left[\frac{(-1{)}^{2N-1}+1‘}{(-1{)}^R}\right]}^{2n-1}C2N-R$

$\left[\frac{-1+1}{(-1{)}^r}\right][as,(-1{)}^{2r-1}]=-1$

$=0$