If the coefficients of x^2 and x^3 are both z | vedclass

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(C) The expansion of $(1 - 3x)^{15}$ is given by $\sum_{k=0}^{15} \binom{15}{k} (-3x)^k = 1 + 15(-3x) + \binom{15}{2}(-3x)^2 + \binom{15}{3}(-3x)^3 +

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$(28, 315)$

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(C) The expansion of $(1 - 3x)^{15}$ is given by $\sum_{k=0}^{15} \binom{15}{k} (-3x)^k = 1 + 15(-3x) + \binom{15}{2}(-3x)^2 + \binom{15}{3}(-3x)^3 + \dots = 1 - 45x + 945x^2 - 13608x^3 + \dots$The expression is $(1 + ax + bx^2)(1 - 45x + 945x^2 - 13608x^3 + \dots)$.The coefficient of $x^2$ is $1(945) + a(-45) + b(1) = 945 - 45a + b = 0$,so $45a - b = 945$ $(1)$.The coefficient of $x^3$ is $1(-13608) + a(945) + b(-45) = -13608 + 945a - 45b = 0$,so $945a - 45b = 13608$. Dividing by $9$,we get $105a - 5b = 1512$ $(2)$.From $(1)$,$b = 45a - 945$. Substituting into $(2)$:$105a - 5(45a - 945) = 1512$$105a - 225a + 4725 = 1512$$-120a = -3213$. This suggests a re-evaluation of the binomial expansion coefficients.Correcting the expansion: $\binom{15}{2} = 105$,$105 	imes 9 = 945$. $\binom{15}{3} = 455$,$455 	imes (-27) = -12285$.Coefficient of $x^2$: $945 - 45a + b = 0 \Rightarrow 45a - b = 945$.Coefficient of $x^3$: $-12285 + 945a - 45b = 0$ $\Rightarrow 945a - 45b = 12285$ $\Rightarrow 21a - b = 273$.Subtracting: $(45a - b) - (21a - b) = 945 - 273$ $\Rightarrow 24a = 672$ $\Rightarrow a = 28$.Substituting $a=28$: $21(28) - b = 273$ $\Rightarrow 588 - b = 273$ $\Rightarrow b = 315$.

If the coefficients of $x^2$ and $x^3$ are both zero in the expansion of the expression $(1 + ax + bx^2)(1 - 3x)^{15}$ in powers of $x$,then the ordered pair $(a, b)$ is equal to

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