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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