The sum of the coefficients of the first thre | vedclass

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(A) The coefficients of the first three terms of $(x - \frac{3}{x^2})^m$ are $^mC_0, -3(^mC_1)$,and $9(^mC_2)$.Given the condition: $^mC_0 - 3(^mC_1)

Vedclass · Accepted Answer

$-5940$

Vedclass · Answer

(A) The coefficients of the first three terms of $(x - \frac{3}{x^2})^m$ are $^mC_0, -3(^mC_1)$,and $9(^mC_2)$.Given the condition: $^mC_0 - 3(^mC_1) + 9(^mC_2) = 559$.Substituting the values: $1 - 3m + \frac{9m(m-1)}{2} = 559$.Multiplying by $2$: $2 - 6m + 9m^2 - 9m = 1118$.$9m^2 - 15m - 1116 = 0$.Dividing by $3$: $3m^2 - 5m - 372 = 0$.Solving the quadratic equation: $(m - 12)(3m + 31) = 0$.Since $m$ is a natural number,$m = 12$.The general term is $T_{r+1} = ^{12}C_r x^{12-r} (-3)^r x^{-2r} = ^{12}C_r (-3)^r x^{12-3r}$.For the term containing $x^3$,set $12 - 3r = 3$,which gives $r = 3$.The required term is $^{12}C_3 (-3)^3 x^3 = 220 	imes (-27) x^3 = -5940x^3$.

The sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x^2})^m$,where $x \neq 0$ and $m$ is a natural number,is $559$. Find the term of the expansion containing $x^3$. (in $x^3$)

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