Question

Show that the coefficient of the middle term in the expansion of (1 + x) ²ⁿ is the sum of the coefficients of the two middle terms in the expansion of (1 + x) ^{2n – 1}.

Answer 1

The expansion of (1 + x)²ⁿ contains (2n + 1 ) terms. 

The middle term here is \((\frac{2n}{2} +1)\)th term, i.e., (n + 1)th term.

    

In the expansion of (1 + x)^{2n – 1}, (2n – 1) being odd, there are two middle terms \((\frac{2n-1+1}{2} )\)th term and \((\frac{2n-1+3}{2} )\)th term, i.e., nth term and (n + 1)th term

∴ In case of (1 + x)^{2n – 1}

t_n = ^{2n – 1}C_{n – 1} x^{n – 1}, t_{n + 1} = ^{2n – 1}C_n xⁿ

^∴ Sum of coefficients of t _n and t _{n + 1} 

= ^{2n – 1}C_{n – 1} + ^{2n – 1}C_n = ^{2n – 1 + 1}C_n = ²ⁿC_n (∵ ⁿC_r + ⁿC_{r + 1} = ^{n + 1}C_{r + 1})

= \(\frac{2n!}{(n!)^2}\) = coefficient of middle term of (1 + x)²ⁿ.