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

Answer»

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}.


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