Question

The greatest term in the expansion of ${(2x+5y)}^{13}$ when $x=10$ and $y=2$ is:

• A. ${{}^{13}C}_5.{20}^7.{10}^4$
• B. ${{}^{13}C}_5.{20}^8.{10}^5$
• C. ${{}^{13}C}_9.{20}^9.{10}^4$
• D. None of these

Answer 1

The correct option is C ${{}^{13}C}_9.{20}^9.{10}^4$

General term, $T_{rH}{=}^{13}{C}_r\left(2n{)}^r\right(5y{)}^{13-r}$

Now, $x=10$ & $y=2$, $\Rightarrow T_{rH}{=}^{13}{C}_r\left(20{)}^r\right(10{)}^{13-r}$

${=}^{13}{C}_r\left(10{)}^{13}\right(2{)}^r$

$\frac{T_{rH}}{T_r}=\frac{{}^{13}{C}_r{.10}^{13}{.2}^r}{{}^{13}{C}_{r-1}{.10}^{13}{.2}^{r-1}}=\frac{2\times 13!}{r.(13-r)!}\times \frac{(r-1)(13-r+1)!}{13!}=\frac{2(13-r+1)}{r}$

$=\frac{26-2r+2}{r}>1=3r<28\Rightarrow 28/3\Rightarrow r<9.33\Rightarrow r=9$

${\Rightarrow }^{13}{C}_9{.10}^{13}{.2}^9{=}^{13}{C}_9{.10}^{13}{2}^9{=}^{13}{C}_9{.20}^9{.10}^4$.