Question

If in the expansion of $(1+x{)}^{20}$, the coefficients of $r^{th}$ and $(r+4{)}^{th}$ terms are equal, then r is equal to?

Answer 1

We are given In the expansion of $(1+x{)}^{20}$, the coefficients of $r^{th}$ and $(r+4{)}^{th}$ terms are equal.

here the any $r^{th}$ term will be

$Tr=\left(\frac{n}{r-1}\right)\left(a{)}^{n-r+1}\right(b{)}^{r-1}$

now, here $a=1$ & $b=x$

so, the coefficient of $r^{th}$ term is $=\left(\frac{n}{r-1}\right)(1{)}^{n-r+1},n=20$

$=\left(\frac{20}{r-1}\right)$

& the co-efficient of $(r+4{)}^{th}$ term is $=\left(\frac{n}{r+4-1}\right)(1{)}^{n-r-4+1},n=20$

$=\left(\frac{20}{r+3}\right)$

so, $\left(\frac{20}{r-1}\right)=\left(\frac{20}{r+3}\right)$

now, it $\left(\frac{n}{r_1}\right)=\left(\frac{n}{r_2}\right)\Rightarrow n=r_1+r_2$

so, $20=r-1+r+3$

$2r+2=20$

$2r=18$

$r=9$

The $r$ is equal to $9$