If the coefficients of x^3 and x^4 in the exp | vedclass

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(B) The expansion is $(1 + ax + bx^2)(1 - 2x)^{18}$.Using the binomial expansion $(1 - 2x)^{18} = \sum_{k=0}^{18} \binom{18}{k} (-2x)^k = \sum_{k=0}^{

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$(16, \frac{272}{3})$

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(B) The expansion is $(1 + ax + bx^2)(1 - 2x)^{18}$.Using the binomial expansion $(1 - 2x)^{18} = \sum_{k=0}^{18} \binom{18}{k} (-2x)^k = \sum_{k=0}^{18} \binom{18}{k} (-2)^k x^k$.Coefficient of $x^n$ in $(1 + ax + bx^2)(1 - 2x)^{18}$ is given by $\binom{18}{n}(-2)^n + a \cdot \binom{18}{n-1}(-2)^{n-1} + b \cdot \binom{18}{n-2}(-2)^{n-2}$.For $x^3$ $(n=3)$: $\binom{18}{3}(-2)^3 + a \cdot \binom{18}{2}(-2)^2 + b \cdot \binom{18}{1}(-2)^1 = 0$.$-8 \cdot \frac{18 \cdot 17 \cdot 16}{6} + 4a \cdot \frac{18 \cdot 17}{2} - 2b \cdot 18 = 0$.$-8 \cdot 816 + 4a \cdot 153 - 36b = 0 \implies -6528 + 612a - 36b = 0 \implies 153a - 9b = 1632 \implies 51a - 3b = 544 \dots (i)$.For $x^4$ $(n=4)$: $\binom{18}{4}(-2)^4 + a \cdot \binom{18}{3}(-2)^3 + b \cdot \binom{18}{2}(-2)^2 = 0$.$16 \cdot \frac{18 \cdot 17 \cdot 16 \cdot 15}{24} - 8a \cdot \frac{18 \cdot 17 \cdot 16}{6} + 4b \cdot \frac{18 \cdot 17}{2} = 0$.$16 \cdot 3060 - 8a \cdot 816 + 4b \cdot 153 = 0 \implies 48960 - 6528a + 612b = 0$.Dividing by $12$: $4080 - 544a + 51b = 0 \implies 544a - 51b = 4080 \dots (ii)$.Solving $(i)$ and $(ii)$: From $(i)$,$3b = 51a - 544 \implies b = 17a - \frac{544}{3}$.Substitute into $(ii)$: $544a - 51(17a - \frac{544}{3}) = 4080 \implies 544a - 867a + 9248 = 4080$.$-323a = -5168 \implies a = 16$.Then $b = 17(16) - \frac{544}{3} = 272 - \frac{544}{3} = \frac{816 - 544}{3} = \frac{272}{3}$.

If the coefficients of $x^3$ and $x^4$ in the expansion of $(1 + ax + bx^2)(1 - 2x)^{18}$ in powers of $x$ are both zero,then $(a, b)$ is equal to

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