Find a if the `17^(t h)`and `18^(t h)`terms of the expansion `(2+a)^(50)`are equal.

Explanation

We know that
General term of expansion `(a+b)^n` is
`T_(r+1)= ^nC_r a^(n-r) b^r " " " " " ....(1)`

Finding `17^(th)` term
`T_(17)=T_(16+1)` of `(2+a)^(50)`
Putting `r=16`, `n=50`,`a=2` and `b=a` in `(1)`
`T_(16+1)= ^(50)C_(16)(2)^(50-16).(a)^(16)`
`T_(17)= ^(50)C_(16)(2)^(34).(a)^(16)`

Finding `18^(th)` term
`T_(18)=T_(17+1)` of `(2+a)^(50)`
Putting `r=17`, `n=50`,`a=2` and `b=a` in `(1)`
`T_(17+1)= ^(50)C_(17)(2)^(50-17).(a)^(17)`
`T_(18)= ^(50)C_(17)(2)^(33).(a)^(17)`

Now it is given that

` " " " " 17^(th)` term=`18^(th)` term

` " " " " ^(50)C_(16)(2)^(34).(a)^(16)` `=^(50)C_(17)(2)^(33).(a)^(17)`

` " " " " ( ^(50)C_(16)(2)^(34))/( ^(50)C_(17)(2)^(33))=((a)^(17))/((a)^(16))`

` " " " " " ``((50!)/(16!(50-16)!))/((50!)/(17!(50-17)!)).(2)^(34-33)=a`

` " " " " " ``((50!)/(16!xx34!))xx((17!xx 33!)/(50!))xx2=a`

` " " " " " ` `(17!xx 33!)/(16!xx34!)xx2=a`

` " " " " " ` `(17xx16!xx 33!)/(16!xx34xx33!)xx2=a`

` " " " " " ` `(17)/(34)xx2=a`

` " " " " " ` `a=(17)/(17)`

` " " " " " ` `a=1`

Hence, `a=1`