I am stuck on a line in a problem. How do I get from
$4-2^{j+1}+4-2^{j-1}$
to
$8-5\times2^{j-1}$
please.
Your help would be greatly appreciated. What technique is it please so I can learn it?
Thanks
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1$2^{j + 1} = 2 \times 2^j$. Now what if you do this sort of thing again? – Joppy May 16 '17 at 12:30
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Hint: write $2^{j+1}=2^{j-1+2}=2^2 \cdot 2^{j-1}$. – Joe May 16 '17 at 12:30
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This is really simple rewriting an exponential. More high school maths. – quallenjäger May 16 '17 at 12:34
2 Answers
1
By the law of exponent addition,
$$2^{j+1}+2^{j-1}=2^j2^1+2^j2^{-1}=2^j(2^1+2^{-1}).$$
You can also write
$$2^{j+1}+2^{j-1}=2^{j-1}2^2+2^{j-1}2^0=2^{j-1}(2^2+1).$$
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$4-2^{j+1}+4-2^{j-1}=8-2^{j+1}-2^{j-1}=8-2^{j-1+2}-2^{j-1}=8-4\cdot2^{j-1}-2^{j-1}=8-5\cdot2^{j-1}$
Jean G
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