Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{t}_{\mathrm{r}}$ is the rth term in the expansion of $(1+\mathrm{x})^{101}$, then what is
the ratio $\frac{\mathrm{t}_{20}}{\mathrm{t}_{19}}$ equal to ?
Options:
the ratio $\frac{\mathrm{t}_{20}}{\mathrm{t}_{19}}$ equal to ?
Solution:
1113 Upvotes
Verified Answer
The correct answer is:
$\frac{83 x}{19}$
We find $r_{n}$ term:
$\mathrm{t}_{\mathrm{r}}$ is the rth term in the expansion of $(1+\mathrm{x})^{101}$. $\mathrm{t}_{\mathrm{r}}=101 \mathrm{C}_{\mathrm{r}-1} \cdot(\mathrm{x})^{(\mathrm{r}-1)}$
$\therefore \frac{\mathrm{t}_{20}}{\mathrm{t}_{19}}=\frac{101}{{ }^{01} \mathrm{C}_{19}} \cdot \frac{\mathrm{x}^{19}}{\mathrm{x}^{18}}=\frac{101}{{ }^{101} \mathrm{C}_{19} \mathrm{x}}{\mathrm{C}_{18}}=\frac{\frac{101 !}{19 ! 82 !}}{\frac{101 !}{18 ! 83 !}} \mathrm{x}=\frac{83 \mathrm{x}}{19}$
$\mathrm{t}_{\mathrm{r}}$ is the rth term in the expansion of $(1+\mathrm{x})^{101}$. $\mathrm{t}_{\mathrm{r}}=101 \mathrm{C}_{\mathrm{r}-1} \cdot(\mathrm{x})^{(\mathrm{r}-1)}$
$\therefore \frac{\mathrm{t}_{20}}{\mathrm{t}_{19}}=\frac{101}{{ }^{01} \mathrm{C}_{19}} \cdot \frac{\mathrm{x}^{19}}{\mathrm{x}^{18}}=\frac{101}{{ }^{101} \mathrm{C}_{19} \mathrm{x}}{\mathrm{C}_{18}}=\frac{\frac{101 !}{19 ! 82 !}}{\frac{101 !}{18 ! 83 !}} \mathrm{x}=\frac{83 \mathrm{x}}{19}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.