The sum of the coefficients in the first three terms of the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m$ is equal to $97$. Find the term of the expansion containing $x^4$.
In the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m$,
first term $= T_1 = C^m_0 \; \left(x^2\right)^m = C^m_0 \times x^{2m}$
second term $= T_2 = C^m_1 \; \left(x^2\right)^{m-1} \; \left(\dfrac{-2}{x}\right) = -2 C^m_1 \times x^{2m-3}$
third term $= T_3 = C^m_2 \; \left(x^2\right)^{m-2} \; \left(\dfrac{-2}{x}\right)^2 = 4 \; C^m_2 \times x^{2m-6}$
$\therefore \;$ Coefficients in the first three terms of the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m$ are $\;$ $C^m_0, \; -2 C^m_1, \; 4C^m_2$.
As per question, $\;\;\;$ $C^m_0 - 2 C^m_1 + 4 C^m_2 = 97$
i.e. $\;$ $1 - \dfrac{2 \times m!}{1! \left(m-1\right)!} + \dfrac{4 \times m!}{2! \left(m-2\right)!} = 97$
i.e. $\;$ $-2m + \dfrac{4m \left(m-1\right)}{2} = 96$
i.e. $\;$ $2m^2 - 4m = 96$
i.e. $\;$ $m^2 - 2m - 48 = 0$
i.e. $\;$ $\left(m-8\right) \left(m+6\right) = 0$
i.e. $\;$ $m = 8$ $\;$ or $\;$ $m = -6$
$\because \;$ $m$ cannot be negative $\implies$ $m = 8$
Let $\left(r+1\right)^{th}$ term in the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m = \left(x^2 - \dfrac{2}{x}\right)^8$ $\;$ be the term of the expansion containing $x^4$.
Now, $\;$ $\left(r+1\right)^{th}$ term $= T_{r+1} = C^8_r \; \left(x^2\right)^{8-r} \; \left(\dfrac{-2}{x}\right)^r$
i.e. $\;$ $T_{r+1} = \left(-2\right)^r \; C^8_r \; x^{16-3r}$
$\therefore \;$ As per question, $\;$ $16 - 3r = 4$
i.e. $\;$ $3r = 12$ $\implies$ $r = 4$
$\therefore \;$ The term containing $x^4$ in the expansion of $\left(x^2 - \dfrac{2}{x}\right)^8$ is
$T_{4+1} = T_5 = \left(-2\right)^4 \; C^8_4 \; x^4$
i.e. $\;$ $T_5 = 16 \times \dfrac{8!}{4! \left(8-4\right)!} \times x^4$
i.e. $\;$ $T_5 = \dfrac{16 \times 8 \times 7 \times 6 \times 5 \times x^4}{4 \times 3 \times 2 \times 1} = 1120 x^4$