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Exercise 6.1 · Q8

Q.Prove that n!(n+2)=n!+(n+1)!n! (n+2) = n! + (n+1)!

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Prove the factorial identity n!(n+2)=n!+(n+1)!n!(n+2) = n! + (n+1)! by expanding (n+1)!(n+1)! in terms of n!n! and factoring the right-hand side.

The defining recursive relation of factorials:

(n+1)!=(n+1)×n!(n+1)! = (n+1)\times n!

which follows directly from n!=n×(n−1)×⋯×1n! = n\times(n-1)\times\cdots\times1, since (n+1)!(n+1)! just has one extra factor of (n+1)(n+1) multiplied in front.

  1. Start from the right-hand side of the identity to be proved: RHS=n!+(n+1)!\text{RHS} = n! + (n+1)!.
  2. Rewrite (n+1)!(n+1)! using the recursive relation: (n+1)!=(n+1) n!(n+1)! = (n+1)\,n!.
  3. Substitute: RHS=n!+(n+1) n!\text{RHS} = n! + (n+1)\,n!.
  4. Both terms share the common factor n!n!; factor it out: RHS=n![1+(n+1)]\text{RHS} = n!\big[1 + (n+1)\big].
  5. Simplify the bracket: 1+(n+1)=n+21 + (n+1) = n+2. …

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