Show that it the greatest term in the expansion of `(1 +x)^(2n)` has also the greatest coefficient, then x lies between and `n/(n +1) and (n+1)/n`.

Explanation

The number of terms in the expansion of `(1+x)^(2n)=2n+1`
`implies`middle term`=(2n+1+1)/(2)=(n+1)^(th)` term
`T_(n+1)` in `(1+x)^(2n)=.^(2n)C_(n)x^(n)`
From the given condition, `T_(n+1)=`middle term=the greatest term
`impliesT_(n+1)gtT_(n) and T_(n+1)gtT_(n+2)`
Now `T_(n+1) gt T_(n) implies.^(2n)C_(n)x^(n)gt.^(2n)C_(n-1)x^(n-1)`
`implies(2n!)/(n!n!)xxgt(2n!)/((n-1)!xx(n+1)!)`
`impliesx gt (n)/(n+1)` . . . (1)
Also `T_(n+1)gtT_(n+2)`
`implies.^(2n)C_(n)x^(n)gt.^(2n)C_(n+1)x^(n+1)`
`implies(2n!)/(n!n!)gt(2n!)/((n+1)!(n-1)!)x`
`implies(n+1)/(n) gt x`
`therefore(n)/(n+1)ltxlt(n+1)/(n)impliex in ((n)/(n+1),(n+1)/(n))`
We know that the greatest term can also be equal to one of the adjancent terms. hence, the equality cann also holds with the inequality (1) & (2) consequently `x in [(n)/(n+1),(n+1)/(n)]`