What is the eighth term given the sequence: +896, -448, +224,-112,....... ?

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What is the eighth term given the sequence: +896, -448, +224,-112,....... ?

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Answer 1 · 2016-09-12T20:04:28

the eighth term is $- 7$ $" "-7$

Explanation:

Let any position count be $i$ $i$
Let any term be $a_{i}$ $a_i$

So
$a_{1} \to first term$ $a_1->" first term"$
$a_{2} \to second term$ $a_2->" second term"$
$a_{i} \to i t h term$ $a_i->" "ith" term"$

Two things I notice:
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Point 1: \to$ $color(blue)("Point 1:")->$ Alternates between positive and negative starting at positive.

So this can be achieved by

$a_{i} \to a_{1} \times {(- 1)}^{2}$ $a_i->a_1 xx(-1)^2$

$a_{i} \to a_{2} \times {(- 1)}^{3}$ $a_i->a_2xx(-1)^3$

$a_{i} \to a_{3} \times {(- 1)}^{4}$ $a_i->a_3xx(-1)^4$

$\Rightarrow a_{i} \to a_{i} \times {(- 1)}^{i + 1}$ $=>a_i->a_ixx(-1)^(i+1)$

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$Point 2: \to$ $color(blue)("Point 2:")->$ Each term is half the previous

So we have

${(- 1)}^{2} \times a_{1}$ $(-1)^2xxa_1$
$a_{2} = {(- 1)}^{3} \times \frac{1}{2} \times a_{1}$ $a_2=(-1)^3xx1/2xxa_1$
$a_{3} = {(- 1)}^{4} \times \frac{1}{2} \times \frac{1}{2} \times a_{1}$ $a_3=(-1)^4xx1/2xx1/2xxa_1$

This implies $a_{i} = {(- 1)}^{i + 1} \times {(\frac{1}{2})}^{i - 1} \times a_{1}$ $a_i=(-1)^(i+1)xx(1/2)^(i-1)xxa_1$

Where $a_{1} = + 896$ $a_1=+896$
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$a_{i} = 896 {(- 1)}^{i + 1} {(\frac{1}{2})}^{i - 1}$ $color(blue)(a_i=896(-1)^(i+1)(1/2)^(i-1)$

Thus

$" "color(purple)(bar(ul(|color(white)(2/2)a_8=896(-1)^(8+1)(1/2)^(8-1) = -7color(white)(2/2)|)))$

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What is the eighth term given the sequence: +896, -448, +224,-112,....... ?

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