Binomial Theorem and its Simple Applications Previous Year Questions

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The coefficient of $x^{- 6}$ , in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ , is __________ . [2023]

Ans.
(5040)

$T_{r + 1} = {}^{9}C_{r} {(\frac{4 x}{5})}^{9 - r} {(\frac{5}{2 x^{2}})}^{r}$

$= {}^{9}C_{r} {(\frac{4}{5})}^{9 - r} {(\frac{5}{2})}^{r} \cdot x^{9 - r - 2 r} \Rightarrow x^{9 - r - 2 r} = x^{- 6}$

$\Rightarrow 9 - r - 2 r = - 6 \Rightarrow 3 r = 15 \Rightarrow r = 5$

$So, T_{6} = {}^{9}C_{5} {(\frac{4}{5})}^{4} {(\frac{5}{2})}^{5} x^{- 6}$

$= \frac{9!}{5! 4!} (\frac{4 \times 4 \times 4 \times 4}{5 \times 5 \times 5 \times 5}) (\frac{5 \times 5 \times 5 \times 5 \times 5}{2 \times 2 \times 2 \times 2 \times 2}) x^{- 6}$

$= \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2} (8) (5) x^{- 6} = 126 \times 40 \cdot x^{- 6} = 5040$

Want Us to Email you About Special Offers & Updates?