suppose that : E= (√( 9 + 4 ( Σ_(k=2) ^n a_( k) ^( 2) ))) is given let a_( k) = a_(k−1) . (a_( k−1) +1) and a_1 =1 , a_( 2) = 2 , a_( 3) = 6 , a_4 = 42 , a_( 5) = 1806 and etc find the value of E =?


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 198813 by mnjuly1970 last updated on 24/Oct/23

$s u p p o s e t h a t :$ $E = \sqrt{9 + 4 (\underset{k = 2}{\sum^{n}} a_{k}^{2})}$ $i s g i v e n$ $l e t a_{k} = a_{k - 1} . (a_{k - 1} + 1)$ $a n d a_{1} = 1, a_{2} = 2, a_{3} = 6,$ $a_{4} = 42, a_{5} = 1806 a n d e t c$ $f i n d t h e v a l u e o f E = ?$
	Answered by witcher3 last updated on 24/Oct/23

	$a_{k} - a_{k - 1} = a_{k - 1}^{2}$ $\Leftrightarrow \underset{k = 3}{\sum^{n + 1}} (a_{k} - a_{k - 1}) = \underset{k = 2}{\sum^{n}} a_{k}^{2}$ $\Leftrightarrow a_{n + 1} - a_{2} = \underset{k = 2}{\sum^{n}} {(a_{k})}^{2}$ $E = \sqrt{9 + 4 (a_{n + 1} - a_{2})} = \sqrt{9 + 4 (a_{n + 1} - 2)}$ $= \sqrt{1 + {4 a}_{n + 1}} = \sqrt{1 + {4 a}_{n} (a_{n} + 1)} = \sqrt{{({2 a}_{n} + 1)}^{2}}$ $=∣ {2 a}_{n} + 1 ∣= {2 a}_{n} + 1, a_{n} > 0$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum