In-a-sequence-if-r-th-term-is-given-by-T-r-2-T-r-1-1-then-give-it-s-n-th-term-in-terms-of-it-s-1-st-term-Given-T-1-2-

Question Number 56074 by Kunal12588 last updated on 10/Mar/19

I n a s e q u e n c e i f r^{t h} t e r m i s g i v e n b y

T_{r} = 2 \times T_{r - 1} + 1

t h e n g i v e {i t}^{'} s n^{t h} t e r m i n t e r m s o f {i t}^{'} s 1^{s t} t e r m

[G i v e n : T_{1} = 2]

Commented by maxmathsup by imad last updated on 10/Mar/19

w e h a v e u_{n} = 2 u_{n - 1} + 1 w i t h u_{1} = 2 l e t f i n d u_{n} i n t e r m s o f n

l e t v_{n} = u_{n} + α l e t d e t r e m i n e α / v_{n} b e a g e o m . s e q u e n c e

v_{n + 1} = u_{n + 1} + α = 2 u_{n} + 1 + α = {q v}_{n} = q (u_{n} + α) \Rightarrow q = 2 a n d 1 + α = q α \Rightarrow

(q - 1) α = 1 \Rightarrow α = 1 \Rightarrow v_{n} = u_{n} + 1 i s g e o m e t r i q u e l e t p r o v e t h a t

v_{n + 1} - v_{n} = u_{n + 1} + 1 = 2 u_{n} + 1 + 1 = 2 (u_{n} + 1) = 2 v_{n} \Rightarrow

v_{n} = v_{1} q^{n - 1} = 3 . 2^{n - 1} \Rightarrow u_{n} = v_{n} - 1 = 3 . 2^{n - 1} - 1 .

Commented by maxmathsup by imad last updated on 10/Mar/19

e r r o r o f t y p o v_{n + 1} = u_{n + 1} + 1

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Mar/19

T_{2} = 2 T_{1} + 1 \to 2 \times 2 + 1 \to 2^{2} + 2^{0}

T_{3} = 2 T_{2} + 1 \to 2 (2^{2} + 1) + 1 \to 2^{3} + 2^{1} + 2^{0}

T_{4} = 2 T_{3} + 1 \to 2 (2^{3} + 2^{1} + 2^{0}) + 1 \to 2^{4} + 2^{2} + 2^{1} + 1

T_{5} = 2 T_{4} + 1 \to 2 (2^{4} + 2^{2} + 2^{1} + 1) + 1 \to 2^{5} + 2^{3} + 2^{2} + 2^{1} + 1

\dots

T_{n} = 2^{n} + 2^{n - 2} + 2^{n - 3} + 2^{n - 4} + \dots + 1 \leftarrow t o t a l n t e r m s

T_{n} = 2^{n} + \frac{2^{n - 2} (1 - \frac{1}{2^{n - 1}})}{(1 - \frac{1}{2})} [s = \frac{a (1 - r^{n})}{(1 - r)}]

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

T_{n} = 2^{n} + 2^{n - 1} - 1 \to i t i s t h e a n s w e r

Commented by Kunal12588 last updated on 10/Mar/19

y e s s i r i t i s c o r r e c t

2^{n - 1} \times 2 + 2^{n - 1} - 1 = 3 \times 2^{n - 1} - 1

Commented by Kunal12588 last updated on 10/Mar/19

t h a n k y o u

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Mar/19

t h a n k y o u s i r \dots

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

l e t T_{n} = {A p}^{n} + C

T_{n} = 2 T_{n - 1} + 1

{A p}^{n} + C - 2 {A p}^{n - 1} - 2 C - 1 = 0

A (p - 2) p^{n - 1} - (C + 1) = 0

\Rightarrow p = 2

\Rightarrow C = - 1

\Rightarrow T_{n} = A \times 2^{n} - 1

T_{1} = A \times 2 - 1 = 2

\Rightarrow A = \frac{3}{2}

\Rightarrow T_{n} = 3 \times 2^{n - 1} - 1

i f t h e q u e s t i o n i s T_{n} = 3 T_{n - 1} + 1

{A p}^{n} + C - 3 {A p}^{n - 1} - 3 C - 1 = 0

A (p - 3) p^{n - 1} - (2 C + 1) = 0

\Rightarrow p = 3

\Rightarrow C = - \frac{1}{2}

\Rightarrow T_{n} = A \times 3^{n} - \frac{1}{2}

\Rightarrow T_{1} = A \times 3 - \frac{1}{2} = 2

\Rightarrow A = \frac{5}{6}

\Rightarrow T_{n} = \frac{1}{2} (5 \times 3^{n - 1} - 1)

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