# Greatest Term of a binomial expansion:

#### Finding the Numerically Greatest term in the binomial expansion:

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.