Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 144767 by mnjuly1970 last updated on 29/Jun/21

	Answered by Rasheed.Sindhi last updated on 29/Jun/21

	${(x^{3} - \frac{2}{x^{2}})}^{m} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{n} x^{n}$ $\frac{a_{0} + a_{1} + a_{2} + . . . + a_{n}}{n + 1} = 0.2 = \frac{1}{5}$ $L e t x = 1$ ${(- 1)}^{m} = a_{0} + a_{1} + a_{2} + . . . + a_{n} = \frac{n + 1}{5}$ $m \in O \to \frac{n + 1}{5} = - 1 \Rightarrow n = - 6 (r e j e c t e d)$ $m \notin O$ $m \in E \to \frac{n + 1}{5} = 1 \Rightarrow n = 4$ $a_{0} + a_{1} + a_{2} + a_{3} + a_{4} = \frac{4 + 1}{5} = 1$ $. . . . . . .$ $. . . . .$
	Answered by mr W last updated on 29/Jun/21

	${(x^{3} - \frac{2}{x^{2}})}^{m} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{n} x^{n}$ $x = 1 :$ ${(1 - \frac{2}{1})}^{m} = a_{0} + a_{1} + a_{2} + . . . + a_{n} = \frac{n + 1}{5}$ ${(- 1)}^{m} = \frac{n + 1}{5}$ $\Rightarrow n = 4, m = 4$ ${(x^{3} - \frac{2}{x^{2}})}^{4} = \frac{16}{x^{8}} - \frac{32}{x^{3}} + 24 x^{2} - 8 x^{7} + x^{12}$ $a_{2} = 24$
	Commented by Rasheed.Sindhi last updated on 30/Jun/21

	$N \overset{∙}{∣} \subset \in Sir!$
	Commented by mnjuly1970 last updated on 30/Jun/21

	$t h x m r W . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum