Binomial Theorem and its Simple Application
119562
The coefficient of \(x^{32}\) in the expansion of \(\left(\mathrm{x}^4-\frac{1}{\mathrm{x}^3}\right)^{15}\) is :
1 \({ }^{-15} \mathrm{C}_3\)
2 \({ }^{15} \mathrm{C}_4\)
3 \({ }^{-15} \mathrm{C}_5\)
4 \({ }^{15} \mathrm{C}_2\)
Explanation:
B The given expansion is \(\left(\mathrm{x}^4-\frac{1}{\mathrm{x}^3}\right)^{15}\)
By the expanding
\({ }^{15} C_0\left(x^4\right)^{15}\left(\frac{-1}{x^3}\right)^0+{ }^{15} C_1\left(x^4\right)^{14}\left(\frac{-1}{x^3}\right)+{ }^{15} C_2\left(x^4\right)^{13}\left(\frac{-1}{x^3}\right)^2+\ldots\)
\(T_{r+1}={ }^{15} C_r\left(x^4\right)^{15-r} \cdot\left(\frac{-1}{x^3}\right)^r\)
\(=-{ }^{15} C_r x^{60-7 r}\)
For the coefficient of \(\mathrm{x}^{32}\)
\(\Rightarrow x^{60-7 \mathrm{r}}=\mathrm{x}^{32}\)
\(\Rightarrow 60-7 \mathrm{r}=32\)
\(\Rightarrow 7 \mathrm{r}=28\)
\(\Rightarrow \mathrm{r}=4\)
\(\begin{aligned} \text { So, } 5^{\text {th }} \text { term, coefficient of } \mathrm{x}^{32} \text { is, } \\ ={ }^{15} \mathrm{C}_4\left(\mathrm{x}^4\right)^{11}\left(-\frac{1}{\mathrm{x}^3}\right)^4={ }^{15} \mathrm{C}_4 \mathrm{x}^{44} \mathrm{x}^{-12} \\ ={ }^{15} \mathrm{C}_4 \mathrm{x}^{32} \\ \text { Thus, coefficient of } \mathrm{x}^{32}={ }^{15} \mathrm{C}_4\end{aligned}\)
