7.Binomial Theorem
hard

Let $\left( a + bx + cx ^2\right)^{10}=\sum \limits_{ i =0}^{20} p _{ i } x ^{ i }, a , b , c \in N$. If $p _1=20$ and $p _2=210$, then $2( a + b + c )$ is equal to

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

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