if-a-1-1-and-a-n-1-3a-n-n-2-find-a-n-

Question Number 76207 by mr W last updated on 25/Dec/19

i f a_{1} = 1 a n d a_{n + 1} = 3 a_{n} + n^{2}

f i n d a_{n} = ?

Answered by mind is power last updated on 25/Dec/19

a_{n + 1} = {3 a}_{n} + n^{2}

a_{n} = {f 3}^{n - 1} + c_{n}

c_{n + 1} = {3 c}_{n} + n^{2}

c_{n} = {an}^{2} + bn + t

a {(n + 1)}^{2} + b (n + 1) + t = {3 an}^{2} + n^{2} + 3 bn + 3 t

a = 3 a + 1

2 a + b = 3 b

a + b = 2 t

a = - \frac{1}{2}, b = - \frac{1}{2}

t = - \frac{1}{2}

a_{n} = {f 3}^{n - 1} - \frac{1}{2} (n^{2} + n + 1)

a_{1} = f - \frac{3}{2} = 1 \Rightarrow f = \frac{5}{2}

a_{n} = \frac{1}{2} (5 . 3^{n - 1} - n^{2} - n - 1)

Answered by john santuy last updated on 25/Dec/19

a_{2} = 3.1 + 1 = 4

a_{3} = 3 \times 4 + 2^{2} = 16

a_{4} = 3 \times 16 + 3^{2} = 57

e t c \dots .

Commented by mr W last updated on 25/Dec/19

i g o t

a_{n} = \frac{5 \times 3^{n - 1} - n^{2} - n - 1}{2}

Commented by john santuy last updated on 25/Dec/19

h o w s i r ?

Answered by mr W last updated on 25/Dec/19

m y a t t e m p t :

a_{n + 2} = 3 a_{n + 1} + {(n + 1)}^{2}

a_{n + 1} = 3 a_{n} + n^{2}

\Rightarrow a_{n + 2} - a_{n + 1} = 3 (a_{n + 1} - a_{n}) + 2 n + 1

l e t b_{n + 2} = a_{n + 2} - a_{n + 1}

\Rightarrow b_{n + 2} = 3 b_{n + 1} + 2 n + 1

a s s u m e b_{n} = c_{n} + p n + q w i t h p, q a s c o n s t a n t s

c_{n + 2} + p (n + 2) + q = 3 [c_{n + 1} + p (n + 1) + q] + 2 n + 1

\Rightarrow c_{n + 2} = 3 c_{n + 1} + 2 (p + 1) n + (p + 2 q + 1)

n o w w e s e l e c t p a n d q s u c h t h a t t h e

c o l o r e d t e r m s b e c o m e z e r o :

p + 1 = 0 \Rightarrow p = - 1

p + 2 q + 1 = 0 \Rightarrow - 1 + 2 q + 1 = 0 \Rightarrow q = 0

i . e . b_{n} = c_{n} - n o r c_{n} = b_{n} + n

f r o m c_{n + 2} = 3 c_{n + 1} w e s e e c_{n} i s a G . P .

w i t h c o m m o n r a t i o 3 .

\Rightarrow c_{n} = 3^{n - 2} c_{2}

a_{1} = 1

a_{2} = 3 \times 1 + 1^{2} = 4

b_{2} = a_{2} - a_{1} = 4 - 1 = 3

c_{2} = b_{2} + 2 = 3 + 2 = 5

\Rightarrow c_{n} = 5 \times 3^{n - 2}

\Rightarrow b_{n} = 5 \times 3^{n - 2} - n

\Rightarrow b_{k + 1} = 5 \times 3^{k - 1} - k - 1

\Rightarrow a_{k + 1} - a_{k} = 5 \times 3^{k - 1} - k - 1

\Rightarrow \underset{k = 1}{\sum^{n}} a_{k + 1} - \underset{k = 1}{\sum^{n}} a_{k} = 5 \times \underset{k = 1}{\sum^{n}} 3^{k - 1} - \underset{k = 1}{\sum^{n}} k - n

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

\Rightarrow a_{n + 1} = \frac{5 (3^{n} - 1) - n (n + 1) - 2 (n - 1)}{2}

o r

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

\Rightarrow a_{n} = \frac{5 \times 3^{n - 1} - n^{2} - n - 1}{2}

Commented by john santuy last updated on 25/Dec/19

w a w w w g r e a t s i r

Answered by vishalbhardwaj last updated on 25/Dec/19

a_{n} = a_{n + 1} - a_{n}

= 3 a_{n} + n^{2} - 3 a_{n - 1} - {(n - 1)}^{2}

= 3 [a_{n} - a_{n - 1}] + n^{2} - n^{2} + 2 n - 1 (i)

= 3 \times c o m m o n d i f f e r e n c e + 2 n - 1

now by the given a_{n + 1} = 3 a_{n} + n^{2}

a_{2} = 3 a_{1} + 1 and a_{2} = 3 (1) + 1 = 4

\Rightarrow common difference = a_{2} - a_{1} = 3

\Rightarrow by equation (i), a_{n} = 3 (3) + 2 n - 1

\Rightarrow a_{n} = 2 n + 8

this is the solution when series an A . P .

Commented by MJS last updated on 25/Dec/19

but obviously it is not an A . P .

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