Question - Question on Sequence Terms from Binomial Expansion and Line Equations

For the binomial expansion part, to find the 4th term from the end in the expansion of $\left( \frac{3}{2} - \frac{x^3}{6} \right)^7$, note that the 4th term from the end is equivalent to the 4th term from the beginning (or the $T_4$ term) because the binomial is symmetric.

To find $T_4$, we use the binomial theorem which states $T_{k+1} = ^nC_k \cdot a^{n-k} \cdot b^{k}$.

$T_4 = ^7C_3 \left( \frac{3}{2} \right)^{7-3} \left( -\frac{x^3}{6} \right)^3$

$T_4 = 35 \cdot \left( \frac{3}{2} \right)^4 \cdot \left( -\frac{x^3}{6} \right)^3$

$T_4 = 35 \cdot \frac{81}{16} \cdot -\frac{x^9}{216}$

$T_4 = -\frac{35 \cdot 81 \cdot x^9}{16 \cdot 216}$

$T_4 = -\frac{35 \cdot 81 \cdot x^9}{6^3 \cdot 6}$

$T_4 = -\frac{35 \cdot 81 x^9}{6^4}$

$T_4 = -\frac{945 \cdot x^9}{1296}$

For the equation of lines passing through (1,2) and making an angle $30^\circ$ with the y-axis, the slope of the line is the tangent of the angle with the x-axis. Since the line makes a $30^\circ$ angle with the y-axis, it makes a $60^\circ$ angle with the x-axis. Hence, the slope $m$ is $\tan(60^\circ) = \sqrt{3}$.

The equation of a line in slope-intercept form is $y = mx + b$.

To find $b$, substitute $(1,2)$ (x, y) into the equation:

$2 = \sqrt{3}(1) + b$

$b = 2 - \sqrt{3}$

Therefore, the equation of the line is:

$y = \sqrt{3}x + (2 - \sqrt{3})$