Math Question and Answers


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 186884 by Mingma last updated on 11/Feb/23

Commented by mr W last updated on 11/Feb/23

$y o u a r e a s k i n g q u e s t i o n s n o n - s t o p,$ $b u t y o u s e e m n e v e r t o g i v e a n y f e e d$ $b a c k : n o t h a n k s t o p e o p l e w h o$ $a n s w e r e d y o u r q u e s t i o n s, n o$ $c o m m e n t s i f t h e a n s w e r s r e a l l y$ $h a v e h e l p e d y o u e t c . . .$ $y o u h a v e p o s t e d t h i s q u e s t i o n i n$ $Q 186657, a n d i t i s a n s w e r e d . {w h a t}^{'} s$ $t h e r e a s o n t h a t y o u p o s t e d t h e s a m e$ $q u e s t i o n h e r e a g a i n ?$
Commented by Mingma last updated on 11/Feb/23
Bro, I give thumbs up when questions are answered. All questions you answered i gave thumbs up.
Commented by Mingma last updated on 11/Feb/23
Bro, I mistakenly posted it.
Commented by mr W last updated on 11/Feb/23

$t h a n k s f o r t h i s f e e d b a c k!$
Commented by Mingma last updated on 11/Feb/23
Thank you too!
Commented by mr W last updated on 11/Feb/23

$i p r e f e r f e e d b a c k s i n w o r d s .$ $‘ ‘ {l i k e s}^{″} a r e n o r e a l c o m m e n t s,$ $t h e y a r e a n o n y m a n d a r e o f t e n$ $m i s u s e d b y s o m e p e o p l e .$
	Answered by ARUNG_Brandon_MBU last updated on 11/Feb/23

	$* a_{1} = 7, d = 8 \Rightarrow a_{n} = 8 n - 1$ $* T_{1} = 3, T_{n + 1} - T_{n} = a_{n}$ $\Rightarrow \underset{k = 1}{\sum^{n}} (T_{k + 1} - T_{k}) = \underset{k = 1}{\sum^{n}} a_{k}$ $\Rightarrow T_{n + 1} - T_{1} = 4 n (n + 1) - n = 4 n^{2} + 3 n$ $\Rightarrow T_{n + 1} = 4 n^{2} + 3 n + 3 \Rightarrow T_{n} = 4 n^{2} - 5 n + 4$ $(i) \underset{k = 1}{\sum^{n}} T_{k} = \frac{2 n (n + 1) (2 n + 1)}{3} - \frac{5 n (n + 1)}{2} + 4 n$ $= \frac{4 n^{3}}{3} + 2 n^{2} + \frac{2 n}{3} - \frac{5 n^{2}}{2} - \frac{5 n}{2} + 4 n$ $= \frac{4 n^{3}}{3} - \frac{n^{2}}{2} + \frac{13 n}{6}$ $(i i) \underset{k = 1}{\sum^{n}} (T_{n + m} - T_{k}) = {n T}_{n + m} - (\frac{4 n^{3}}{3} - \frac{n^{2}}{2} + \frac{13 n}{6})$ $= 4 n {(n + m)}^{2} - 5 n (n + m) + 4 n - (\frac{4 n^{3}}{3} - \frac{n^{2}}{2} + \frac{13 n}{6})$ $= \frac{8 n^{3}}{3} + 8 n^{2} m + 4 {n m}^{2} - \frac{9 n^{2}}{2} - 5 n m + \frac{11 n}{6}$ $(i i i) 3 T_{n + 2} - 3 T_{n + 1} + T_{n}$ $= 3 (4 {(n + 2)}^{2} - 5 (n + 2) + 4) - 3 (4 {(n + 1)}^{2} - 5 (n + 1) + 4) + 4 n^{2} - 5 n + 4$ $= 3 (4 n^{2} + 11 n + 10) - 3 (4 n^{2} + 3 n + 3) + 4 n^{2} - 5 n + 4$ $= 4 n^{2} + 19 n + 25$
	Commented by Mingma last updated on 11/Feb/23
	Great work!
	Commented by Mingma last updated on 12/Feb/23
	I like the effort and demonstrations of the results. I wonder if a reduction of the third part could be done to bring the solution into a form recognizable in terms of T_{p}, where p = ?.
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum