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Induction: sum of n(n+1)(n+2)

Short answer

Basis n=1: 1·2·3 = 1·2·3·4/4. Assume sum to k = k(k+1)(k+2)(k+3)/4. Add (k+1)(k+2)(k+3), factor (k+1)(k+2)(k+3) to get (k+1)(k+2)(k+3)(k+4)/4. True for all n≥1.

Prove by Mathematical Induction: 1.2.3 + 2.3.4 + 3.4.5 + ........ + n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4

To prove. P(n): 1·2·3 + 2·3·4 + … + n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4 for all n ≥ 1.

Basis step (n = 1)

  • LHS = 1·2·3 = 6.
  • RHS = 1·2·3·4/4 = 6.
  • P(1) is true.

Inductive hypothesis

Assume P(k): 1·2·3 + … + k(k+1)(k+2) = k(k+1)(k+2)(k+3)/4.

Inductive step

Show P(k+1): the sum up to (k+1)(k+2)(k+3) equals (k+1)(k+2)(k+3)(k+4)/4.

LHS = k(k+1)(k+2)(k+3)/4 + (k+1)(k+2)(k+3)      (hypothesis)
    = (k+1)(k+2)(k+3) [ k/4 + 1 ]
    = (k+1)(k+2)(k+3) (k+4)/4  = RHS

Conclusion

By mathematical induction, P(n) is true for every positive integer n. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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