Question

The numerically greatest term inthe expansion (2x-3y)^12 when x=2 nd y=5/2​

Answer 1

To find: We have to find the numerically greatest term of $(2x-3y)^{12}$ when $x=2,\:y=\frac{5}{2}$.

Solution:

Here the general term,

$\quad t_{r+1}=\binom{12}{r}.(2x)^{12-r}.(-3y)^r$

$\therefore t_{r}=\binom{12}{r-1}.(2x)^{12-r+1}.(-3y)^{r-1}$

$\therefore\left|\frac{t_{r+1}}{t_{r}}\right|=\frac{\binom{12}{r}}{\binom{12}{r-1}}.\frac{(2x)^{12-r}.(3y)^{r}}{(2x)^{12-r+1}.(3y)^{r-1}}$

$\quad\quad=\frac{13-r}{r}.\frac{3y}{2x}$

$\quad\quad=\frac{13-r}{r}.\frac{\frac{15}{2}}{4}\quad[\because x=2,\:y=\frac{5}{2}]$

$\quad\quad=\frac{15}{8}.\big(\frac{13}{r}-1\big)$

$\therefore \frac{|t_{r+1}|}{|t_{r}|}\gtreqqless 1$ according as $\frac{15}{8}\big(\frac{13}{r}-1\big)\gtreqqless 1$

i.e., according as $\frac{13}{r}-1\gtreqqless \frac{8}{15}$

i.e., according as $\frac{13}{r}\gtreqqless \frac{23}{15}$

i.e., according as $\frac{1}{r}\gtreqqless \frac{23}{195}$

i.e., according as $r\lesseqqgtr\frac{195}{23}\approx 8.5$

Now, $[8.5]=8$

So, $|t_{r+1}|\gtreqqless |t_{r}|$ according as $r\lesseqqgtr 8$.

Hence, $t_{8+1}=t_{9}$ is the greatest numerical term, so that the greatest numerical term is

$\quad=\binom{12}{8}.(2x)^{12-8}.(3y)^{8}$

$\quad=\binom{12}{8}.(4)^{4}.\big(\frac{15}{2}\big)^{8}\quad[\because x=2,\:y=\frac{5}{2}]$

$\quad=\frac{12!}{8!\:(12-8)!}.2^{8}.\frac{15^{8}}{2^{8}}$

$\quad=\frac{12.11.10.9.8!}{8!.4!}.15^{8}$

$\quad=\frac{4.3.11.5.2.3.3}{4.3.2}.15^{8}$

$\quad=11.5.3.3.15^{8}$

$\quad=\bold{1268630859375}$

Answer: The numerically greatest term in the expansion of $(2x-3y)^{12}$ is $\bold{1268630859375}$ when $x=2,\:y=\frac{5}{2}$.

Answer 2

Given:

(2x-3y)¹² , x=1 and y= 5/2  

To find:

The numerically greatest term in the expansion (2x-3y)¹² when x=1 and y= 5/2 is​

Solution:

From given, we have,

(2x-3y)¹², x=1 and y= 5/2  

we use the formula,

P = [ (n + 1) |x| ] / [|x| + 1]  

to find the numerically greatest term.

so we have,

(2x-3y)¹² = (2x)¹²(1 - 3y/2x)¹²

Thus we get,

|x| = (3/2) (y/x)

substituting the values of x and y, we get,

|x| = (3/2) (5/2) = 15/4

Now consider,

P = [ (n + 1)|x| ] / [|x| + 1]  

P = [ (12 + 1) (15/4)] / [15/4 + 1]  

P = [ (13) (15/4)] / [19/4]  

P = 195/19 = 10.26

Thus 10.26 ≈ 11, 11th term is the numerically greatest term in the expansion (2x-3y)¹² when  x=1 and y= 5/2.