Question

If x⁴ occurs in the r^th terms in the expansion of ${\left(x^4+\frac{1}{x^3}\right)}^{15},$ then r = ___________.

Answer 1

$\mathrm{for}{\left(x^4+\frac{1}{x^3}\right)}^{15}$
n = 15
$\begin{array}{rcl}{T}_{r+1}& =& {}^{15}C_4\left(x^{4r}\right){\left(\frac{1}{x^3}\right)}^{15-r}\\ & =& {}^{15}C_4{x}^{4r}{x}^{-\left(45-3r\right)}\\ & =& {}^{15}C_4{x}^{4r}{x}^{3r-45}\\ & =& {}^{15}C_4{x}^{7r-45}\end{array}$
To get x⁴, 7r – 45 = 4
i.e. 7r = 49
i.e r = 7