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Exercises · Q4

Q.Prove the following by using the principle of mathematical induction for all n∈Nn \in N: 1.2.3+2.3.4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)41.2.3 + 2.3.4 + \ldots + n(n+1)(n+2) = \dfrac{n(n+1)(n+2)(n+3)}{4}

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Let P(n)P(n) be the statement

1⋅2⋅3+2⋅3⋅4+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)4.1\cdot2\cdot3+2\cdot3\cdot4+\ldots+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}.

Base case: For n=1n=1,

LHS=1⋅2⋅3=6,RHS=1⋅2⋅3⋅44=244=6.\text{LHS}=1\cdot2\cdot3=6,\qquad \text{RHS}=\frac{1\cdot2\cdot3\cdot4}{4}=\frac{24}{4}=6.

So P(1)P(1) is true.

Inductive step: Assume P(k)P(k) is true for some k≥1k\ge1:

1⋅2⋅3+…+k(k+1)(k+2)=k(k+1)(k+2)(k+3)4.(Induction Hypothesis)1\cdot2\cdot3+\ldots+k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4}. \qquad \text{(Induction Hypothesis)}

We must show the sum up to (k+1)(k+1) equals (k+1)(k+2)(k+3)(k+4)4\dfrac{(k+1)(k+2)(k+3)(k+4)}{4}.

Adding the next term (k+1)(k+2)(k+3)(k+1)(k+2)(k+3) to both sides of the hypothesis: …

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