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 22768 by math solver last updated on 22/Oct/17

Commented by math solver last updated on 22/Oct/17

$f i n d t h e s u m o f i n f i n i t e t e r m s o f$ $t h e s e r i e s . .$
Commented by math solver last updated on 22/Oct/17

$s o m e o n e s o l v e t h i s o n e p l z .$
	Answered by $@ty@m last updated on 22/Oct/17
	$N^r of t_n =2n+1 D^r of t_n =1^2 +2^2 +...+n^2 =Σn^2 =((n(n+1)(2n+1))/6) (you can find its proof in your textbook) ∴ t_n =((6(2n+1))/(n(n+1)(2n+1))) =(6/(n(n+1))) =6((1/n)−(1/(n+1))) ∴S_n =6{(1−(1/2))+((1/2)−(1/3))+.....+((1/n)−(1/(n+1)))} =6(1−(1/(n+1))) =((6n)/(n+1)) S_∞ =lim_(n→∞) ((6n)/(n+1)) S_∞ =lim_(n→∞) (6/(1+(1/n)))=6 Ans.$
	$N^{r} o f t_{n} = 2 n + 1$ $D^{r} o f t_{n} = 1^{2} + 2^{2} + . . . + n^{2}$ $= Σ n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ $(y o u c a n f i n d i t s p r o o f i n y o u r t e x t b o o k)$ $∴ t_{n} = \frac{6 (2 n + 1)}{n (n + 1) (2 n + 1)}$ $= \frac{6}{n (n + 1)}$ $= 6 (\frac{1}{n} - \frac{1}{n + 1})$ $∴ S_{n} = 6 {(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + . . . . . + (\frac{1}{n} - \frac{1}{n + 1})}$ $= 6 (1 - \frac{1}{n + 1})$ $= \frac{6 n}{n + 1}$ $S_{\infty} = \lim_{n \to \infty} \frac{6 n}{n + 1}$ $S_{\infty} = \lim_{n \to \infty} \frac{6}{1 + \frac{1}{n}} = 6 A n s .$
	Commented by math solver last updated on 22/Oct/17

	$h o w y o u w r i t e t_{n} d i r e c t l y ?$ $p l z t e l l s t e p s ? ? i a l w a y s h a v e p r o b l e m$ $f i n d i n g t h a t . . . . . .$
	Commented by $@ty@m last updated on 22/Oct/17

	$I s i t c l e a r n o w ?$
	Commented by math solver last updated on 22/Oct/17

	$y e s, t h a n k s b r o!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum