Alaaalaaalaa
Alaaalaaalaa
19-08-2015
Mathematics
contestada
n(n+1)(n+2)(n+3)(n+4)
Respuesta :
Аноним
Аноним
19-08-2015
n(n+1)(n+2)(n+3)(n+4)
=(n(n+1)(n+2)(n+3))(n+4)
=(n(n+1)(n+2)(n+3))(n)+(n(n+1)(n+2)(n+3))(4)
=n^5+6n^4+11n^3+6n^2+4n^4+24n^3+44n^2+24n
=n^5+10n^4+35n^3+50n^2+24n
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