If the coefficients of `(r-5)^(th)` and `(2r-1)^(t h)`terms of the expansion `(1+x)^(34)` are equal, find r.

Explanation

We know that

General term of expansion `(a+b)^n` is

` " " " ``T_(r+1)=^nC_r " a^(n-r) b^r`

General term for `(1+x)^(34)`

Putting `a=1`, `b=x`, `n=34`

` " " " ``T_(r+1)=^(34)C_r " (1)^(n-r) x^r`

` " " " ``T_(r+1)=^(34)C_r " x^r " " " " .....(1)`

Coefficient of `(r-5)^(th)` term

i.e. `T_(r-5)`

i.e. `T_(r-6+1)`

Putting `r=r-6` in `(1)`

`T_(r-6+1)=^34C_(r-6)(x)^(r-6)`

`T_(r-5)=^34C_(r-6)(x)^(r-6)`

`therefore` Coefficient of `(r-5)^(th)` term= ` " ^34C_(r-6)`


Coefficient of `(2r-1)^(th)` term

i.e. `T_(2r-1)`

i.e. `T_(2r-2+1)`

Putting `r=2r-2` in `(1)`

`T_(2r-2+1)=^34C_(2r-2)(x)^(2r-2)`

`T_(2r-1)=^34C_(2r-2)(x)^(2r-2)`

`therefore` Coefficient of `(2r-1)^(th)` term= ` " ^34C_(2r-2)`

Given that,

Coefficient of `(r-5)^(th)` & `(2r-1)^(th)` term are equal

i.e. ` " ^34C_(r-6)=^34C_(2r-2)`

We know that if `^nC_(r)=^nC_(p)` then `r=p` or `r=n-p`

So,

`r-6=2r-2` ` " " " ` or ` " " " ` `r-6=34-(2r-2)`


`r-6=2r-2`
`-6+2=2r-r`
`r=-4`


`r-6=34-(2r-2)`
`r+2r=34+2+6`
`3r=42`
`r=14`

Since `r` is a natural number

So `r=-4` not possible

Hence, `r=14`