riverea338338
riverea338338
25-05-2018
Mathematics
contestada
h(n)=n^3+2n; Find h(n+4)
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gmany
gmany
27-05-2018
[tex]h(n)=n^3+2n\\\\h(n+4)=(n+4)^3+2(n+4)=n^3+3n^2\cdot4+3n\cdot4^2+4^3+2n+2\cdot4\\\\=n^3+12n^2+48n+64+2n+8=n^3+12n^2+50n+72[/tex]
[tex]Used:\\\\(a+b)^3=a^3+3a^2b+3ab^2+b^3\\\\a\cdot(b+c)=ab+ac[/tex]
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