Lyric Morris

2022-02-05

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

vhudzeniy2n

Step 1
Let any position count be i
Let any term be ${a}_{i}$
So
${a}_{1}\to$ first term
${a}_{2}\to$ second term
${a}_{i}\to$ its term
Two things I notice:
Step 1
Alternates between positive and negative starting at positive.
So this can be achieved by
${a}_{i}\to {a}_{1}×{\left(-1\right)}^{2}$
${a}_{i}\to {a}_{2}×{\left(-1\right)}^{3}$
${a}_{i}\to {a}_{3}×{\left(-1\right)}^{4}$
$⇒{a}_{i}\to {a}_{i}×{\left(-1\right)}^{i+1}$
Step 2
Each term is half the previous
So we have
${\left(-1\right)}^{2}×{a}_{1}$
${a}_{2}={\left(-1\right)}^{3}×\frac{1}{2}×{a}_{1}$
${a}_{3}={\left(-1\right)}^{4}×\frac{1}{2}×\frac{1}{2}×{a}_{1}$
This implies ${a}_{1}={\left(-1\right)}^{i+1}×{\left(\frac{1}{2}\right)}^{i-1}×{a}_{1}$
Step 3
${a}_{i}=896{\left(-1\right)}^{i+1}{\left(\frac{1}{2}\right)}^{i-1}$
Thus
${a}_{8}=896{\left(-1\right)}^{8+1}{\left(\frac{1}{2}\right)}^{8-1}=-7$

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