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  • 18-08-2015
  • Mathematics
Answered

n(n+1)(n+2)(n+3)(n+4)

Verified Answer :

Anonymous Anonymous
  • 18-08-2015

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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