Let \(k^{th}\) term is \(T_k\) and \((k + 1)^{th}\) term is \(T_{k + 1}\).
we already know that $$ T_k = \binom{n}{k - 1} \ a^{n - k + 1} \ b^{k - 1} $$ $$ = \frac{n!}{(n - k + 1)! (k - 1)!} a^{n - k + 1} b^{k - 1} $$ then \((k + 1)^{th}\) term. $$ T_{k + 1} = \binom{n}{k} \ a^{n - k} \ b^k $$ $$ = \frac{n!}{(n - k)! k!} a^{n - k} b^k $$ hence
$$ \frac{T_{k + 1}}{T_k} = \frac{(n - k + 1)! (k - 1)!}{(n - k)! k!} (a^{(n - k)} b^k) (a^{(- n + k - 1)} b^{(- k + 1)}) $$
\(\frac{T_{k + 1}}{T_k} = \frac{(n - k + 1)! (k - 1)!}{(n - k)! k!} (a^{(n - k)} b^k) \\ (a^{(- n + k - 1)} b^{(- k + 1)})\)
$$ = \frac{(n - k + 1) (n - k + 1 - 1)! (k - 1)!}{(n - k)! k \times (k - 1)!} (a^{-1}) b $$
\( = \frac{(n - k + 1) (n - k + 1 - 1)! (k - 1)!}{(n - k)! k \times (k - 1)!} (a^{-1}) b \)
$$ = \frac{(n - k + 1) (n - k)! (k - 1)!}{(n - k)! k \times (k - 1)!} \ \frac{b}{a} $$
$$ = \frac{(n - k + 1)}{k} \ \frac{b}{a} $$
Hence \(T_{k + 1} \gt T_k\)
If \(\frac{(n - k + 1)}{k} \ \frac{b}{a} \gt 1\) then $$ (n - k + 1) \frac{b}{a} \gt k $$ $$ (n + 1) \frac{b}{a} \gt k + k \frac{b}{a} $$ $$ (n + 1) \frac{b}{a} \gt k (1 + \frac{b}{a}) $$ $$ \frac{(n + 1) \frac{b}{a}}{(1 + \frac{b}{a})} \gt k $$ $$ \frac{(n + 1) \frac{b}{a}}{\frac{b}{a}(\frac{a}{b} + 1)} \gt k $$ $$ \frac{(n + 1)}{(\frac{a}{b} + 1)} \gt k $$ or $$ k \lt \frac{(n + 1)}{(\frac{a}{b} + 1)} $$ Similarly if \(T_{k + 1} \lt T_k\) then. $$ k \gt \frac{(n + 1)}{(\frac{a}{b} + 1)} $$ Similarly if \(T_{k + 1} = T_k\) then. $$ k = \frac{(n + 1)}{(\frac{a}{b} + 1)} $$
Case(1): If the value of \(\frac{(n + 1)}{(\frac{a}{b} + 1)}\) is a positive integer P, then the \(P^{th}\) term and \(( p + 1)^{th}\) term will be the numerically greatest terms of the binomial expansion \((a + b)^n\).
Case(2): If the value of \(\frac{(n + 1)}{(\frac{a}{b} + 1)}\) is in the form of (P + Q), where P is a positive integer and Q is a rational number \((0 \lt Q \lt 1)\), then the \(( p + 1)^{th}\) term will be the numerically greatest term of the binomial expansion \((a + b)^n\).
Note: While finding the numerically greatest term, the negative sign is to be neglected.
Example(1): Find the numerically greatest term in the binomial expansion of \((3 - 2x)^9\), where \(x = \frac{1}{2}\)?
Solution: Given, \(n = 9\), \(a = 3\), \(b = 2x\), \(x = \frac{1}{2}\) then the value of \(\frac{(n + 1)}{(\frac{a}{b} + 1)}\)
$$ = \frac{(9 + 1)}{(\frac{3}{2x} + 1)} $$ $$ = \frac{(9 + 1)}{(\frac{3}{2 \frac{1}{2}} + 1)} $$ $$ = \frac{10}{3 + 1} $$ $$ = \frac{10}{4} $$ $$ = 2.5 $$ Here the value of \(\frac{(n + 1)}{(\frac{a}{b} + 1)}\) is in the form of (P + Q), where P = 2 and Q = 0.5. Hence only \(2^{nd}\) term will be the numerically greatest term.
Example(2): Find the numerically greatest term in the binomial expansion of \((2x + 3y)^{12}\), where \(x = 3\) and \(y = 2\)?
Solution: Given, \(n = 12\), \(a = 2x\), \(b = 3y\), \(x = 3\), and \(y = 2\) then the value of \(\frac{(n + 1)}{(\frac{a}{b} + 1)}\)
$$ = \frac{(12 + 1)}{(\frac{2x}{3y} + 1)} $$ $$ = \frac{13}{\frac{2 \times 3}{3 \times 2} + 1} $$ $$ = \frac{13}{\frac{6}{6} + 1} $$ $$ = \frac{13}{1 + 1} $$ $$ = \frac{13}{2} $$ $$ = 6.5 $$ Here the value of \(\frac{(n + 1)}{(\frac{a}{b} + 1)}\) is in the form of (P + Q), where P = 6 and Q = 0.5. Hence only \(6^{th}\) term will be the numerically greatest term.
Example(3): Find the numerically greatest term in the binomial expansion of \((1 + 3x)^{11}\), where \(x = \frac{2}{3}\)?
Solution: Given, \(n = 11\), \(a = 1\), \(b = 3x\), and \(x = \frac{2}{3}\), then the value of \(\frac{(n + 1)}{(\frac{a}{b} + 1)}\)
$$ = \frac{(11 + 1)}{(\frac{1}{3x} + 1)} $$ $$ = \frac{12}{\frac{1}{3 \times \frac{2}{3}} + 1} $$ $$ = \frac{12}{\frac{1}{2} + 1} $$ $$ = \frac{12}{\frac{3}{2}} $$ $$ = \frac{24}{3} $$ $$ = 8 $$ Here the value of \(\frac{(n + 1)}{(\frac{a}{b} + 1)}\) is a positive integer P. Hence \(8^{th}\) term and \(9^{th}\) term will be the numerically greatest terms.