By mathematical induction $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\dots +\frac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}$

detailed solution

Correct option is B

Let P(n):11⋅2⋅3+12⋅3⋅4+…+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)Step l: For n=1LHS=11.2.3=16 and RHS =1(1+3)4(1+1)(1+2)=16,P(1 ) is true.Step II: Let P(k) is true, thenP(k):11⋅2⋅3+12⋅3⋅4+…+1k(k+1)(k+2)=k(k+3)4(k+1)(k+2)Step lll : For n=n+1,thenP(k+1):11⋅2⋅3+12⋅3⋅4+…+1k(k+1)(k+2)+1(k+1)(k+2)(k+3)=(k+1)(k+4)4(k+2)(k+3)  LHS =11⋅2⋅3+12⋅3⋅4+…+1k(k+1)(k+2)=k(k+3)4(k+1)(k+2)+1(k+1)(k+2)(k+3)=k(k+3)2+44(k+1)(k+2)(k+3)=k3+6k2+9k+44(k+1)(k+2)(k+3)=(k+1)2(k+4)4(k+1)(k+2)(k+3)=(k+1)(k+4)4(k+2)(k+3)= RHS Hence, P(k + 1) is true. Hence, by principle of mathematical induction for all n ∈ N, P(n) is true.