7.Binomial Theorem
hard

ધારોકે $\left(a+b x+c x^2\right)^{10}=\sum \limits_{i=0}^{20} p_i x^i a, b, c \in N$ જો $p_1=20$ અને $p_2=210$ હીય, તો $2(a+b+c)=.......$

A

$8$

B

$12$

C

$15$

D

$6$

(JEE MAIN-2023)

Solution

$\left(a+b x+c x^2\right)^{10}=\sum_{i=0}^{20} p_i x^i$

Coefficient of $x^1=20$

$20=\frac{10 !}{9 ! 1 !} \times a^9 \times b^1$

$a^9 . b =2$

$a=1, b=2$

Coefficient of $x ^2=210$

$210=\frac{10 !}{9 ! 1 !} \times a^9 \times c^1+\frac{10 !}{8 ! 2 !} \times a^8 b^2$

$210=10 . c+45 \times 4$

$10 c=30$

$c=3$

$2(a+b=c)=12$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.