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 182512 by mathlove last updated on 10/Dec/22

	Answered by manxsol last updated on 10/Dec/22

	$\sqrt{7}$
	Answered by manxsol last updated on 11/Dec/22

	$a n a l i s i s 11 - 2 a_{1} ⟩ 0 \Rightarrow a_{1} ⟨ 5.5$ $0 ⟨ x ⟨ \sqrt{11}$ $x = \sqrt{11 - 2 a_{1}}$ $a_{1} =^{3} \sqrt{17 - 3 a_{2}}$ $a_{2} =^{4} \sqrt{97 - 4 a_{3}}$ $a_{3} =^{5} \sqrt{1049 - 5 d}$ $. . . . = . . . . . .$ $x^{2} = 11 - 2 a_{1}$ $a_{1}^{3} = 17 - 3 a_{2}$ $a_{2}^{4} = 97 - 4 a_{3}$ $a_{3}^{5} = 1048 - 5 a_{4}$ $. . . . . . . . . . . . . . . . . . . . . . . . . .$ $x^{2} - 7 = 2^{2} - 2 a_{1}$ $a_{1}^{3} = 2^{3} + 3^{2} - 3 a_{2}$ $a_{2}^{4} = 3^{4} + 4^{2} - 4 a_{3}$ $a_{3}^{5} = 4^{5} + 5^{2} - 5 a_{4}$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $x^{2} - 7 = 2 (2 - a_{1})$ $a_{1}^{3} - 2^{3} = 3 (3 - a_{2})$ $a_{2}^{4} - 3^{4} = 4 (4 - a_{3})$ $a_{3}^{5} - 4^{5} = 5 (5 - a_{4})$ $. . . . . . . . . . . . . . . . . . . . . . . . . .$ $x^{2} - 7 = 2 (2 - a_{1})$ $(a_{1} - 2) P_{1} = 3 (3 - a_{2})$ $(a_{2} - 3) P_{2} = 4 (4 - a_{3})$ $(a_{3} - 4) P_{3} = 5 (5 - a_{4})$ $P_{n} = [a_{n}^{n + 2} - {(n + 1)}^{n + 2}] / [a_{n} - (n + 1)]$ $. . . . . . . . . . . . . . . . . . . . .$ $m u l t i p l i c a t i o n$ $(x^{2} - 7) P_{1} P_{2} P_{3} . . . = 2.3.4.5$ $x^{2} - 7 = \frac{n!}{P_{1} P_{2} P_{3 . . . . . .}}$ $x^{2} = 7 + \frac{n!}{∐ P n}$ $n \Rightarrow \infty \Rightarrow \frac{n!}{∐ P_{n}} \Rightarrow 0 ∴ x^{2} = 7$ $x = \sqrt{7}$ $x = \sqrt{7} a p p r o v e s$ $i n i t i a l c o n d i t i o n$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum