Sums and differences of powers (1 Viewer)

Jonothancroller · Oct 1, 2015

Hey, I'm trying to do a question but I keep getting a different answer for the first part:

The answers say that F(n+1)-2=((2^n)+1)((2^n)+1) but when I was doing the question I got F(n+1)-2=((2^n+1)+1)((2^n+1)-1). Can someone show me how to get the answer, thanks.

InteGrand · Oct 1, 2015

Jonothancroller said:
Hey, I'm trying to do a question but I keep getting a different answer for the first part:
View attachment 32478
View attachment 32479
The answers say that F(n+1)-2=((2^n)+1)((2^n)+1) but when I was doing the question I got F(n+1)-2=((2^n+1)+1)((2^n+1)-1). Can someone show me how to get the answer, thanks.

$F_{n+1}-2=2^{2^{n+1}}+1-2$

$=2^{2^{n+1}}-1$

$$=2^{2^{n}\times 2}-1$ (by index laws: $2^{n+1}=2\times 2^{n}$)$

$$=\left(2^{2^n}\right)^2-1^2$ (by index laws: $\left(2^{2^n}\right)^2=2^{2^{n}\times 2}$)$

$$=\left(2^{2^n}+1\right)\left(2^{2^n}-1\right)$ (difference of two squares),$

$as the answers have.$

Jonothancroller · Oct 1, 2015

InteGrand said:
$F_{n+1}-2=2^{2^{n+1}}+1-2$

$=2^{2^{n+1}}-1$

$$=2^{2^{n}\times 2}-1$ (by index laws: $2^{n+1}=2\times 2^{n}$)$

$$=\left(2^{2^n}\right)^2-1^2$ (by index laws: $\left(2^{2^n}\right)^2=2^{2^{n}\times 2}$)$

$$=\left(2^{2^n}+1\right)\left(2^{2^n}-1\right)$ (difference of two squares),$

$as the answers have.$

Oh did the answers mean:
$$=\left(2^{2^n}+1\right)\left(2^{2^n}-1\right)$
when they wrote
$$=\left(2^{n}+1\right)\left(2^{n}-1\right)$
And that's why they did F(n)[F(n)-2)

InteGrand · Oct 1, 2015

Jonothancroller said:
Oh did the answers mean:
$$=\left(2^{2^n}+1\right)\left(2^{2^n}-1\right)$
when they wrote
$$=\left(2^{n}+1\right)\left(2^{n}-1\right)$
And that's why they did F(n)[F(n)-2)

Lol I didn't even notice their typo. Yes, that's what they meant.

Jonothancroller · Oct 1, 2015

InteGrand said:
Lol I didn't even notice their typo. Yes, that's what they meant.

Thanks a lot.

Sums and differences of powers (1 Viewer)

Jonothancroller

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InteGrand

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Jonothancroller

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InteGrand

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Jonothancroller

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