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

Explanation

In the expansion of `(1 + x)^(2n) ` , the middle term is
` ((2n)/(2) + 1) ` th
i.e., `(n+1) ` the term , we know that from binomial expansion ,
middle term has greatest coefficient
`[ because "Terms" T_(1), T_(2), T_(3) ,…, T_(n), T_(n), T_(n+1) T_(n+2),...] `
`therefore T_(n) lt T_(n +1) gt T_(n+ 2)`
`rArr (T_(n+1))/(T_(n) )= (""^(2n)C_(n)*x^(n))/(""^(2n)C_(n-1)= (2n - n + 1)/(n) x `
` rArr (T_(n+2))/(T_(n)) gt 1 "or:" (n+1)/(n) , x gt 1`
or ` x gt (n)/(n+1)` (i)
and `rArr (T_(n+2))/(T_(n+1) )= (""^(2n)C_(n+1)*x^(n=1))/( ""^(2n)C_(n) *x^(6)) = (2n - (n + 1)-3)/(n+1) x= (n)/(n+1) . x `
` rArr (T_(n+2))/(T_(n+1)) gt 1 rArr (n)/(n+2) , x lt 1 " or " x lt (n+1)/(n)` ...(ii)
From Eqs . (i) and (ii) , we get
` (n)/(n +1) lt xlt (n+1)/(n)`
Corollary For n = 5
` (5)/(6) lt x lt (6)/(5)`
and ` a_(1) - a_(3) + a_(5) - ... = sin ((npi)/(2))`
Putting `x = omega " and omega^(2) ` (cube roots of unity in Eq . (i) , we get
`a_(0) + a_(1) omega +a_(2) omega^(2) + a_(3)omega^(2) + a_(4) omega^(4) + ... = 0 `..(iv)
and ` a_(0) + a_(1) omega^(2) + a_(2) omega^(4) + a_(3) omega^(6) + a_(4)omega^(8) + ... = 0 `...(v)
On addomg Eqs. (ii),(iv) and (v) and then dividing by 3 , we get
` a_(0) + a_(3) + a_(6) + ... = 3^(n-1)` .