A-sequence-of-numbers-T-1-T-2-T-3-Satisfies-the-relation-3T-n-T-n-1-6-Show-that-for-all-values-of-n-1-T-n-3-1-3-T-n-3-3-Hence-or-otherwise-show-that-if-T-1-

Question Number 10485 by Saham last updated on 13/Feb/17

A sequence of numbers T_{1}, T_{2}, T_{3}, \dots Satisfies the

relation {3 T}_{n} = T_{n - 1} + 6, Show that for all values

of n ⩾ 1, T_{n} - 3 = \frac{1}{3} (T_{n - 3} - 3) . Hence or otherwise

show that if T_{1} = 4, T_{n} = 3 + 3^{1 - n} for all values of

n, also find the sum of the first five terms of the

sequence .

Answered by mrW1 last updated on 19/Feb/17

{3 T}_{n} = T_{n - 1} + 6 \Rightarrow T_{n - 1} = 3 T_{n} - 6 \dots (i)

{3 T}_{n - 1} = T_{n - 2} + 6 \dots (i i)

{3 T}_{n - 2} = T_{n - 3} + 6 \Rightarrow T_{n - 2} = \frac{T_{n - 3}}{3} + 2 \dots (i i)

(i) a n d (i i i) i n (i i) :

3 (3 T_{n} - 6) = (\frac{T_{n - 3}}{3} + 2) + 6

9 T_{n} = \frac{T_{n - 3}}{3} + 26

T_{1} = 4 = 3 + 1 = 3 + 3^{0}

T_{2} = \frac{1}{3} T_{1} + 2 = \frac{4}{3} + 2 = 3 + 3^{- 1}

T_{3} = \frac{1}{3} T_{2} + 2 = \frac{1}{3} (3 + 3^{- 1}) + 2 = 3 + 3^{- 2}

T_{4} = \frac{1}{3} T_{3} + 2 = \frac{1}{3} (3 + 3^{- 2}) + 2 = 3 + 3^{- 3}

\dots

T_{n} = 3 + 3^{1 - n}

S (n) = \underset{k = 1}{\sum^{n}} T_{k} = \underset{k = 1}{\sum^{n}} (3 + 3^{1 - k}) = 3 n + 1 \times \frac{1 - {(\frac{1}{3})}^{n}}{1 - \frac{1}{3}}

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

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

S (5) = 3 \times 5 + \frac{3^{5} - 1}{2 \times 3^{5 - 1}} = 15 + \frac{243 - 1}{2 \times 81} = 15 + \frac{121}{81} = 16 \frac{40}{81} \approx 16.494

Commented by Saham last updated on 19/Feb/17

God bless you sir . i will reduce it .

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