Discussion
The coefficient of \(x^{8}\) in \((x^{2} + x^{3} + x^{4} + x^{5})^{5}\) is :
Answer: D
The coefficient is 0.
Every term of \((x^{2} + x^{3} + x^{4} + x^{5})^{5}\) is a product of five factors, each of degree at least 2, so every term has degree at least \(5 \times 2 = 10\). In fact
\[ \begin{aligned} (x^{2} + \cdots + x^{5})^{5} \\ = x^{10}(1 + x + x^{2} + x^{3})^{5} \end{aligned} \]
so the lowest power present is \(x^{10}\), and \(x^{8}\) does not occur.
13, 10 and 20 would be coefficients of higher powers; for example the coefficient of \(x^{11}\) is 5. The trap is to count solutions of \(e_{1} + \cdots + e_{5} = 8\) without noticing each \(e_{i} \ge 2\).
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development