- Home
- Standard 11
- Mathematics
7.Binomial Theorem
hard
माना $\left(\mathrm{a}+\mathrm{bx}+\mathrm{cx}^2\right)^{10}=\sum_{\mathrm{i}=0}^{20} \mathrm{p}_{\mathrm{i}} \mathrm{x}^{\mathrm{i}}, \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{N}$ है। यदि $\mathrm{p}_1=20$ तथा $\mathrm{p}_2=210$ हैं, तो $2(\mathrm{a}+\mathrm{b}+\mathrm{c})$ बराबर है :
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
