Mathematics - Sequence & Series Question with Solution | TestHub

We know by Binomial expansion, that $(x+a{)}^{\mathrm{n}}$ $\begin{array}{rl}& ={}^{\mathrm{n}}{\mathrm{C}}_0{x}^{\mathrm{n}}{\mathrm{a}}^0+{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{x}}^{\mathrm{n}-1}\cdot \mathrm{a}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{x}}^{\mathrm{n}-2}{\mathrm{a}}^2\\ & +{}^{\mathrm{n}}{\mathrm{C}}_3{\mathrm{x}}^{\mathrm{n}-3}{\mathrm{a}}^3+{}^{\mathrm{n}}{\mathrm{C}}_4{\mathrm{x}}^{\mathrm{n}-4}\cdot {\mathrm{a}}^4+\dots ..+{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^0{a}^{\mathrm{n}}\end{array}$ Given expansion is ${\left(x^4-\frac{1}{x^3}\right)}^{15}$ On comparing we get $\mathrm{n}=15,\mathrm{x}={\mathrm{x}}^4$, $a=\left(-\frac{1}{x^3}\right)$ $\begin{array}{rl}& \therefore {\left(x^4-\frac{1}{x^3}\right)}^{15}={}^{15}{\mathrm{C}}_0{\left({\mathrm{x}}^4\right)}^{15}{\left(-\frac{1}{{\mathrm{x}}^3}\right)}^0\\ & +{}^{15}{\mathrm{C}}_1{\left({\mathrm{x}}^4\right)}^{14}\left(-\frac{1}{{\mathrm{x}}^3}\right)+{}^{15}{\mathrm{C}}_2{\left({\mathrm{x}}^4\right)}^{13}{\left(-\frac{1}{{\mathrm{x}}^3}\right)}^2\\ & +{}^{15}{\mathrm{C}}_3{\left({\mathrm{x}}^4\right)}^{12}{\left(-\frac{1}{{\mathrm{x}}^3}\right)}^3+{}^{15}{\mathrm{C}}_4{\left({\mathrm{x}}^4\right)}^{11}{\left(-\frac{1}{{\mathrm{x}}^3}\right)}^4+\dots \\ & {\mathrm{T}}_{\mathrm{r}+1}={}^{15}{\mathrm{C}}_{\mathrm{r}}{\left({\mathrm{x}}^4\right)}^{15-\mathrm{r}}\cdot {\left(-\frac{1}{{\mathrm{x}}^3}\right)}^{\mathrm{r}}=-{}^{15}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{60-7\mathrm{r}}\\ & \Rightarrow \mathrm{x}{}^{60-7\mathrm{r}}={\mathrm{x}}^{32}\Rightarrow 60-7\mathrm{r}=32\\ & \Rightarrow 7\mathrm{r}=28\Rightarrow \mathrm{r}=4\end{array}$ So, 5 th term, contains ${\mathrm{x}}^{32}$ $={}^{15}{C}_4\left(x^4\right){}^{11}{\left(-\frac{1}{x^3}\right)}^4={}^{15}{C}_4{x}^{44}{x}^{-12}$ $={}^{15}{\mathrm{C}}_4{\mathrm{x}}^{32}.$ Thus, coefficient of ${\mathrm{x}}^{32}={}^{15}{\mathrm{C}}_4$.