If the sum of the coefficients of the first,s | vedclass

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(A) The expansion is $(x^2 + \frac{1}{x})^m$. The first three terms are $^mC_0(x^2)^m$,$^mC_1(x^2)^{m-1}(\frac{1}{x})$,and $^mC_2(x^2)^{m-2}(\frac{1}{

Vedclass · Accepted Answer

$84$

Vedclass · Answer

(A) The expansion is $(x^2 + \frac{1}{x})^m$. The first three terms are $^mC_0(x^2)^m$,$^mC_1(x^2)^{m-1}(\frac{1}{x})$,and $^mC_2(x^2)^{m-2}(\frac{1}{x})^2$.Since the coefficients are $^mC_0$,$^mC_1$,and $^mC_2$,we have $^mC_0 + ^mC_1 + ^mC_2 = 46$.$1 + m + \frac{m(m-1)}{2} = 46$.$2 + 2m + m^2 - m = 92$.$m^2 + m - 90 = 0$.$(m+10)(m-9) = 0$. Since $m > 0$,$m = 9$.The general term is $T_{r+1} = ^9C_r(x^2)^{9-r}(x^{-1})^r = ^9C_r x^{18-2r-r} = ^9C_r x^{18-3r}$.For the term independent of $x$,$18 - 3r = 0 \Rightarrow r = 6$.The coefficient is $^9C_6 = ^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.

If the sum of the coefficients of the first,second,and third terms of the expansion of $(x^2 + \frac{1}{x})^m$ is $46$,then the coefficient of the term that does not contain $x$ is:

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