If , then is equal to [2025]

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.
If $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$ , then ${(α + β)}^{2}$ is equal to [2025]

1 27

2 18

3 81

4 9

Ans.
(3)

We have, $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$

Now, $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r}$

$= \sum_{r = 1}^{9} (\frac{r}{2^{r}}) \cdot {}^{9}C_{r} + \sum_{r = 1}^{9} (\frac{3}{2^{r}}) \cdot {}^{9}C_{r}$

$= \sum_{r = 1}^{9} (\frac{9}{2^{r}}) \cdot {}^{8}C_{r - 1} + 3 \sum_{r = 1}^{9} {}^{9}C_{r} {(\frac{1}{2})}^{r}$

$= \frac{9}{2} \sum_{r = 1}^{9} {}^{8}C_{r - 1} {(\frac{1}{2})}^{r - 1} + 3 (\sum_{r = 0}^{9} ({}^{9}C_{r} {(\frac{1}{2})}^{r}) - 1)$

$= \frac{9}{2} {(1 + \frac{1}{2})}^{8} + 3 ({(1 + \frac{1}{2})}^{9} - 1)$

$= \frac{9}{2} \cdot {(\frac{3}{2})}^{8} + 3 {(\frac{3}{2})}^{9} - 3$

$= 6 \cdot {(\frac{3}{2})}^{9} - 3$

On comparing, we get

$\Rightarrow α = 6 and β = 3$

$∴ {(α + β)}^{2} = 81$ .

Want Us to Email you About Special Offers & Updates?