If the coefficients of ${(2r+1)}^{\text{th}}$  term and ${(r+2)}^{\text{th}}$  term in the expansion of ${(1+x)}^{49}$  are equal, then $r=$

Given  expansion  ${(1+x)}^{49}$ is,

We have general term in the expansion ${\left(x+a\right)}^n$

$(\therefore T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$ be the expansion of ${(x+a)}^n)$  

${(2r+1)}^{\text{th}}$  term

$T_{2r+1}={}^{49}{\text{C}}_{2r}{\left(1\right)}^{49-2r}{x}^{2r}$  

 $T_{2r+1}{=}^{49}{\text{C}}_{2r}{x}^{2r}$

And ${(r+2)}^{\text{th}}$  term

and $T_{r+2}={}^{49}{\text{C}}_{r+1}{\left(1\right)}^{49-r}{x}^{r+1}$

$T_{r+2}={}^{49}{\text{C}}_{r+1}{x}^{r+1}$

According to problem  the coefficients of ${(2r+1)}^{\text{th}}$  and ${(r+2)}^{\text{th}}$  terms are equal

Given  ${}^{49}{C}_{2r}={}^{49}{C}_{r+1}$ $\left({\therefore }^n{\text{C}}_{2r}{=}^n{\text{C}}_{r-1}\Rightarrow 2r=r-1,n=2r+r-1\right)$

$\Rightarrow 2r=r+1$  or $2r+r+1=49$

$\Rightarrow$                 $r=1$  or  $r=16$