Question

If $x^4$ occurs in the $rth$ term in the expansion of ${(x^4+\frac{1}{x^3})}^{15}$, then $r=$

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is A $7$

${(x^4+\frac{1}{x^3})}^{15}{=}^{15}{C}_0(x^4{)}^0\times {\left(\frac{1}{x^3}\right)}^{15}+........{+}^{15}{C}_r(x^4{)}^r{\left(\frac{1}{x^3}\right)}^{15-r}+.......$

Thus $T_{r+1}{=}^{15}{C}_r{x}^{4r}{\left(\frac{1}{x}\right)}^{45-3r}$

$T_{r+1}{=}^{15}{C}_4{x}^{7r-45}$

to get $x^4\Rightarrow 7r-45=4$

$7r=49$

$r=7$

Thus $T_8{=}^{15}{C}_4{x}^4$