Let

Question:

Let $\left(2 x^{2}+3 x+4\right)^{10}=\sum_{r=0}^{20} a_{r} x^{r}$. Then $\frac{a_{7}}{a_{13}}$ is equal to___________.

Solution:

The given expression is $\left(2 x^{2}+3 x+4\right)^{10}=\sum_{r=0}^{20} a_{r} x^{r}$

General term $=\frac{10 !}{r_{1} ! r_{2} ! r_{3} !}\left(2 x^{2}\right)^{\eta}(3 x)^{r_{2}}(4)^{r_{3}}$

Since, $a_{7}=$ Coeff. of $x^{7}$

$2 r_{1}+r_{2}=7$ and $r_{1}+r_{2}+r_{3}=10$

Possibilities are

$a_{7}=\frac{10 ! 3^{7} 4^{3}}{7 ! 3 !}+\frac{10 !(2)(3)^{5}(4)^{4}}{5 ! 4 !}$

$+\frac{10 !(2)^{2}(3)^{3}(4)^{5}}{2 ! 3 ! 5 !}+\frac{10 !(2)^{3}(3)(4)^{6}}{3 ! 6 !}$

$a_{13}=$ Coeff. of $x^{13}$

$2 r_{1}+r_{2}=13$ and $r_{1}+r_{2}+r_{3}=10$

Possibilities are

$a_{13}=\frac{10 !\left(2^{3}\right)\left(3^{7}\right)}{3 ! 7 !}+\frac{10 !\left(2^{4}\right)\left(3^{5}\right)(4)}{4 ! 5 !}$

$+\frac{10 !\left(2^{5}\right)\left(3^{3}\right)\left(4^{2}\right)}{5 ! 3 ! 2 !}+\frac{10 !\left(2^{6}\right)(3)\left(4^{3}\right)}{6 ! 1 ! 3 !}$

$\therefore \frac{a_{7}}{a_{13}}=2^{3}=8$

 

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