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Question Number 98380 by Rio Michael last updated on 13/Jun/20

let {u_{n}} and {v_{n}} be sequences defined by

u_{0} = 9, u_{n + 1} = \frac{1}{2} u_{n} - 3 .

v_{n} = u_{n} + 6 .

Calculate P_{n} = \underset{i = 0}{\sum^{n}} V_{i} in terms of n, the deduce Q_{n} = \underset{i = 0}{\sum^{n}} u_{i}

using the above expressions find

\lim_{x \to \infty} Q_{n} .

Answered by Aziztisffola last updated on 13/Jun/20

P_{n} = 30 - 15 {(\frac{1}{2})}^{n}

Q_{n} = 2 (1 - {(\frac{1}{2})}^{n + 1}) - 6 (n + 1)

Double subscripts: use braces to clarify

Answered by mr W last updated on 13/Jun/20

u_{n + 1} = \frac{1}{2} u_{n} - 3

u_{n + 1} + 6 = \frac{1}{2} (u_{n} + 6)

\Rightarrow u_{n} + 6 = (u_{0} + 6) {(\frac{1}{2})}^{n} = \frac{15}{2^{n}}

\Rightarrow u_{n} = \frac{15}{2^{n}} - 6

\Rightarrow v_{n} = \frac{15}{2^{n}}

P_{n} = \underset{i = 0}{\sum^{n}} v_{i} = 15 \times \frac{1 - \frac{1}{2^{n + 1}}}{1 - \frac{1}{2}} = 30 (1 - \frac{1}{2^{n + 1}})

Q_{n} = \underset{i = 0}{\sum^{n}} u_{i} = \underset{i = 0}{\sum^{n}} (v_{i} - 6) = 30 (1 - \frac{1}{2^{n + 1}}) - 6 (n + 1)

\lim_{n \to \infty} Q_{n} = - \infty

Commented by Rio Michael last updated on 13/Jun/20

perfect now sir thanks

Commented by Rio Michael last updated on 13/Jun/20

sir that is correct . thanks to you both .

my worry is : why (n + 1) ?

i know Σ V_{i} = Σ u_{i} + Σ 6 \Rightarrow Σ u_{i} = Σ V_{i} - Σ 6

why will it not equal 30 [1 - {(\frac{1}{2})}^{n}] - 6 n instead it equals

30 [1 - {(\frac{1}{2})}^{n + 1}] - 6 n - 6

Commented by mr W last updated on 13/Jun/20

b e c a u s e i = 0 t o n, t o t a l l y n + 1 t i m e s,

a n d P_{0} \neq 0, Q_{0} \neq 0 .

\underset{i = 0}{\sum^{n}} P_{i} \neq \underset{i = 1}{\sum^{n}} P_{i}

\underset{i = 0}{\sum^{n}} Q_{i} \neq \underset{i = 1}{\sum^{n}} Q_{i}

Answered by mathmax by abdo last updated on 13/Jun/20

v_{n} = u_{n} + 6 \Rightarrow v_{n + 1} = u_{n + 1} + 6 = \frac{1}{2} u_{n} - 3 + 6 = \frac{1}{2} u_{n} + 3 = \frac{1}{2} (u_{n} + 6) = \frac{1}{2} v_{n}

\Rightarrow v_{n} is g . prog . \Rightarrow v_{n} = v_{0} {(\frac{1}{2})}^{n} we have v_{0} = u_{0} + 6 = 15 \Rightarrow

v_{n} = \frac{15}{2^{n}} we have P_{n} = \sum_{i = 0}^{n} v_{i} = v_{0} \times \frac{1 - q^{n + 1}}{1 - q} = 15 \times \frac{1 - \frac{1}{2^{n + 1}}}{\frac{1}{2}}

= 15 (2 - \frac{2}{2^{n + 1}}) = 15 (2 - \frac{1}{2^{n}}) \Rightarrow P_{n} = 30 - \frac{15}{2^{n}}

Q_{n} = \sum_{i = 0}^{n} u_{i} = \sum_{i = 0}^{n} (v_{i} - 6) = \sum_{i = 0}^{n} v_{i} - 6 (n + 1)

= 30 - \frac{15}{2^{n}} - 6 (n + 1) \Rightarrow \lim_{n \to + \infty} Q_{n} = - \infty