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 194204 by Rupesh123 last updated on 30/Jun/23

	Answered by MM42 last updated on 30/Jun/23

	$i f n = m = 0 \Rightarrow 2 a_{0} - 1 = \frac{1}{2} (2 a_{0}) \Rightarrow a_{0} = 1$ $i f m = 1, n = 0 \Rightarrow 2 a_{1} - 2 = \frac{1}{2} (a_{2} + a_{0})$ $\Rightarrow 6 - 2 = \frac{1}{2} (a_{2} + 1) \Rightarrow a_{2} = 7$ $m = 2, n = 0 \Rightarrow 2 a_{2} - 3 = \frac{1}{2} (a_{4} + a_{0})$ $11 = \frac{1}{2} (a_{4} + 1) \Rightarrow a_{4} = 21$ $m = 2, n = 1 \Rightarrow a_{3} + a_{1} - 2 = \frac{1}{2} (a_{4} + a_{2})$ $a_{3} + 1 = \frac{1}{2} (21 + 7) \Rightarrow a_{3} = 13$ $m = 3, n = 0 \Rightarrow 2 a_{3} - 4 = \frac{1}{2} (a_{6} + a_{0})$ $44 = a_{6} + 1 \Rightarrow a_{6} = 43$ $m = 3, n = 2 \Rightarrow a_{5} + a_{1} - 2 = \frac{1}{2} (a_{6} + a_{4})$ $a_{5} + 1 = \frac{1}{2} (43 + 21) \Rightarrow a_{5} = 31$ $\Rightarrow a_{n} : 1, 3, 7, 13, 21, 31, 43, . . .$ $\Rightarrow a_{n} = a_{n - 1} + 2 n; n ⩾ 1, a_{0} = 1$ $\Rightarrow a_{n} = 1 + 2 + 4 + . . . + 2 n =$ $1 + 2 (1 + 2 + . . . + n) = n^{2} + n + 1$ $\Rightarrow a_{2024} = 2024^{2} + 2024 + 1 = 4098601 ✓$
	Commented by Rupesh123 last updated on 30/Jun/23
	Perfect ��
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum