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7.Binomial Theorem
hard
If the coefficients of $x^2$ and $x^3$ are both zero, in the expansion of the expression $(1 + ax + bx^2) (1 -3x)^{t5}$ in powers of $x$, then the ordered pair $(a, b)$ is equal to
A
$(-54, 315)$
B
$(28, 861)$
C
$(28, 315)$
D
$(-21, 714)$
(JEE MAIN-2019)
Solution
Coefficient of $x^{2}=^{15} C_{2} \times 9-3 a\left(^{15} C_{1}\right)+b=0$
$\Rightarrow^{15} \mathrm{C}_{2} \times 9-45 \mathrm{a}+\mathrm{b}=0………(1)$
Coefficient of $x^{3}=-27 \times^{15} \mathrm{C}_{3}+9 \mathrm{a} \times^{15} \mathrm{C}_{2}-3 \mathrm{b} \times^{15} \mathrm{C}_{1}=0$
$\Rightarrow-273+21 a-b=0………(2)$
$(1)+(2)$ give
$-24 a+672=0$
$\Rightarrow \quad a=28, b=315$
Standard 11
Mathematics
