How do you evaluate $(a^{n + 2} - 2 a^{n} + 3 a^{n + 1}) (a^{n} + a^{n + 1})$ $(a ^ { n + 2} - 2a ^ { n } + 3a ^ { n + 1} ) ( a ^ { n } + a ^ { n + 1} )$ ?

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How do you evaluate $(a^{n + 2} - 2 a^{n} + 3 a^{n + 1}) (a^{n} + a^{n + 1})$ $(a ^ { n + 2} - 2a ^ { n } + 3a ^ { n + 1} ) ( a ^ { n } + a ^ { n + 1} )$ ?

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Answer 1 · 2017-11-19T00:58:55

$- 2 a^{2 n} + a^{2 n + 1} + 4 a^{2 n + 2} + a^{2 n + 3}$ $-2a^(2n)+a^(2n+1)+4a^(2n+2)+a^(2n+3)$
OR
$a^{2 n} \cdot (- 2 + a + 4 a^{2} + a^{3})$ $a^(2n)*(-2+a+4a^2+a^3)$

Explanation:

Laws:
1) $x^{a} \cdot x^{b} = x^{a + b}$ $x^a*x^b=x^(a+b)$
2) $b \cdot x^{a} + c \cdot x^{a} = (b + c) \cdot x^{a}$ $b*x^a+c*x^a=(b+c)*x^(a)$

$(a^{n + 2} - 2 a^{n} + 3 a^{n + 1}) (a^{n} + a^{n + 1}) =$ $(a^(n+2)-2a^n+3a^(n+1))(a^n+a^(n+1))=$
$= a^{n + 2 + n} + a^{n + 2 + n + 1} - 2 a^{n + n} - 2 a^{n + n + 1} + 3 a^{n + n + 1} + 3 a^{n + 1 + n + 1} =$ $=a^(n+2+n)+a^(n+2+n+1)-2a^(n+n)-2a^(n+n+1)+3a^(n+n+1)+3a^(n+1+n+1)=$
$= a^{2 n + 2} + a^{2 n + 3} - 2 a^{2 n} - 2 a^{2 n + 1} + 3 a^{2 n + 1} + 3 a^{2 n + 2} =$ $=a^(2n+2)+a^(2n+3)-2a^(2n)-2a^(2n+1)+3a^(2n+1)+3a^(2n+2)=$
$= 4 a^{2 n + 2} + a^{2 n + 3} - 2 a^{2 n} + a^{2 n + 1} =$ $=4a^(2n+2)+a^(2n+3)-2a^(2n)+a^(2n+1)=$
$= - 2 a^{2 n} + a^{2 n + 1} + 4 a^{2 n + 2} + a^{2 n + 3} =$ $=-2a^(2n)+a^(2n+1)+4a^(2n+2)+a^(2n+3)=$

$= a^{2 n} \cdot (- 2 + a + 4 a^{2} + a^{3})$ $=a^(2n)*(-2+a+4a^2+a^3)$

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How do you evaluate $(a^{n + 2} - 2 a^{n} + 3 a^{n + 1}) (a^{n} + a^{n + 1})$ $(a ^ { n + 2} - 2a ^ { n } + 3a ^ { n + 1} ) ( a ^ { n } + a ^ { n + 1} )$ ?

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How do you evaluate $(a^{n + 2} - 2 a^{n} + 3 a^{n + 1}) (a^{n} + a^{n + 1})$ $(a ^ { n + 2} - 2a ^ { n } + 3a ^ { n + 1} ) ( a ^ { n } + a ^ { n + 1} )$ ?