If-T-n-1-1-1-2-T-n-Find-a-formular-for-T-n-in-terms-of-n-and-find-the-sum-of-first-n-terms-

Question Number 75327 by TawaTawa last updated on 09/Dec/19

If T_{n + 1} = 1 + \frac{1}{2} T_{n}

Find a formular for T_{n} in terms of n

and find the sum of first n terms

Commented by malwaan last updated on 11/Dec/19

T_{1} = 4

T_{2} = 1 + \frac{1}{2} \times 4 = 1 + 2 = 3

T_{3} = 1 + \frac{1}{2} \times 3 = 1 + \frac{3}{2} = \frac{5}{2}

T_{4} = 1 + \frac{1}{2} \times \frac{5}{2} = 1 + \frac{5}{4} = \frac{9}{4}

T_{5} = 1 + \frac{9}{8} = \frac{17}{8} = \frac{2^{4} + 1}{2^{3}} = \frac{2^{5 - 1} + 1}{2^{5 - 2}}

\Rightarrow T_{n} = \frac{2^{n - 1} + 1}{2^{n - 2}}

Commented by mr W last updated on 10/Dec/19

p l e a s e c h e c k t h e q u e s t i o n :

w h e n g i v i n g a r e c u r s i v e s e q u e n c e,

t h e f i r s t t e r m T_{1} m u s t b e g i v e n i n

o r d e r t o d e t e r m i n e a l l t e r m s .

Commented by TawaTawa last updated on 10/Dec/19

Ohh, yes sir, T_{1} = 4

Please help me edit it in your work . Thanks sir

Answered by mr W last updated on 10/Dec/19

s a y T_{n} = {A p}^{n} + B

T_{n + 1} = {A p}^{n + 1} + B

{A p}^{n + 1} + B = 1 + \frac{{A p}^{n}}{2} + \frac{B}{2}

\Rightarrow {A p}^{n} (p - \frac{1}{2}) + \frac{B}{2} = 1

\Rightarrow p - \frac{1}{2} = 0 \Rightarrow p = \frac{1}{2}

\Rightarrow \frac{B}{2} = 1 \Rightarrow B = 2

\Rightarrow T_{n} = \frac{A}{2^{n}} + 2

i f T_{1} = 4,

4 = \frac{A}{2} + 2

\Rightarrow A = 4

\Rightarrow T_{n} = 2 + \frac{1}{2^{n - 2}}

\underset{k = 1}{\sum^{n}} T_{k} = 2 n + \frac{2 (1 - \frac{1}{2^{n}})}{1 - \frac{1}{2}} = 2 (n + 2) - \frac{1}{2^{n - 2}}

Commented by TawaTawa last updated on 10/Dec/19

God bless you sir . I appreciate your time

Commented by TawaTawa last updated on 10/Dec/19

I really appreciate sir . God bless you more .

Commented by TawaTawa last updated on 10/Dec/19

Sir, is there any hint on how we assume T_{n} = {Ap}^{n} + B .

Or it is general for such questions

Commented by TawaTawa last updated on 10/Dec/19

And please sir, why is p - \frac{1}{2} = 0 and \frac{B}{2} = 1

Commented by mind is power last updated on 10/Dec/19

T_{n + 1} = {aT}_{n} + b

if a = 1 \Leftrightarrow T_{n + 1} - T_{n} = b \Rightarrow \underset{k = 1}{\sum^{n - 1}} (T_{k + 1} - T_{k}) = \underset{k = 1}{\sum^{n - 1}} b

\Rightarrow T_{n} = b (n - 1) + T_{1}

if a \neq 1

let x = ax + b \Rightarrow x = \frac{b}{1 - a}

let

T_{n} = W_{n} + \frac{b}{1 - a}

\Rightarrow W_{n + 1} + \frac{b}{1 - a} = a (W_{n} + \frac{b}{1 - a}) + b

\Rightarrow W_{n + 1} + \frac{ab}{1 - a} = {aW}_{n} + \frac{ab}{1 - a} + b = {aW}_{n} + \frac{b}{1 - a}

\Rightarrow W_{n + 1} = {aW}_{n} \Rightarrow W_{n} = W_{1} {(a)}^{n - 1}

T_{n} = \frac{b}{1 - a} + W_{1} (a^{n - 1})

W_{1} = T_{1} - \frac{b}{1 - a}

Commented by TawaTawa last updated on 10/Dec/19

God bless you sir, thanks for your time