**Question:**

If the coefficients of $(r-5)$ th and $(2 r-1)$ th terms in the expansion of $(1+$$x)^{34}$ are equal, find the value of r.

**Solution:**

To find: the value of r with respect to the binomial expansion of $(1+$$x)^{34}$ where the coefficients of the (r – 5)th and (2r – 1)th terms are equal to each other

Formula Used:

The general term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where

${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$

Now, finding the (r – 5)th term, we get

$T_{r-5}={ }^{34} C_{r-6} \times x^{r-6}$

Thus, the coefficient of $(r-5)$ th term is ${ }^{34} C_{r-6}$

Now, finding the (2r – 1)th term, we get

$T_{2 r-1}={ }^{34} C_{2 r-2} \times(x)^{2 r-2}$

Thus, coefficient of $(2 r-1)$ th term is ${ }^{34} \mathrm{C} 2 r-2$

As the coefficients are equal, we get

${ }^{34} C_{2 r-2}={ }^{34} C_{r-6}$

$2 r-2=r-6$

$R=-4$

Value of $r=-4$ is not possible

$2 r-2+r-6=34$

$3 r=42$

$R=14$

Thus, value of $r$ is 14