Question-186884

Question Number 186884 by Mingma last updated on 11/Feb/23

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

you are asking questions non−stop, but you seem never to give any feed back: no thanks to people who answered your questions, no comments if the answers really have helped you etc... you have posted this question in Q 186657, and it is answered. what′s the reason that you posted the same question here again?

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 \dots

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 prefer feedbacks in words. “likes”are no real comments, they are anonym and are often misused by some people.

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 ⇒a_n =8n−1 ∗T_1 =3, T_(n+1) −T_n =a_n ⇒Σ_(k=1) ^n (T_(k+1) −T_k )=Σ_(k=1) ^n a_k ⇒T_(n+1) −T_1 =4n(n+1)−n=4n^2 +3n ⇒T_(n+1) =4n^2 +3n+3 ⇒T_n =4n^2 −5n+4 (i) Σ_(k=1) ^n T_k =((2n(n+1)(2n+1))/3)−((5n(n+1))/2)+4n =((4n^3 )/3)+2n^2 +((2n)/3)−((5n^2 )/2)−((5n)/2)+4n =((4n^3 )/3)−(n^2 /2)+((13n)/6) (ii) Σ_(k=1) ^n (T_(n+m) −T_k )=nT_(n+m) −(((4n^3 )/3)−(n^2 /2)+((13n)/6)) =4n(n+m)^2 −5n(n+m)+4n−(((4n^3 )/3)−(n^2 /2)+((13n)/6)) =((8n^3 )/3)+8n^2 m+4nm^2 −((9n^2 )/2)−5nm+((11n)/6) (iii)3T_(n+2) −3T_(n+1) +T_n =3(4(n+2)^2 −5(n+2)+4)−3(4(n+1)^2 −5(n+1)+4)+4n^2 −5n+4 =3(4n^2 +11n+10)−3(4n^2 +3n+3)+4n^2 −5n+4 =4n^2 +19n+25

* 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 = ?.

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