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Exercises · Q14

Q.Prove the following by using the principle of mathematical induction for all n∈Nn \in N: (1+11)(1+12)(1+13)…(1+1n)=(n+1)\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\ldots\left(1+\dfrac{1}{n}\right) = (n+1)

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For a positive integer rr, the general factor is

1+1r=r+1r.1+\frac1r=\frac{r+1}{r}.

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

(1+11)(1+12)⋯(1+1n)=n+1.\left(1+\frac11\right)\left(1+\frac12\right)\cdots\left(1+\frac1n\right)=n+1.

Base case: For n=1n=1,

LHS=1+11=2,RHS=1+1=2.\text{LHS}=1+\frac11=2,\qquad \text{RHS}=1+1=2.

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

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

(1+11)⋯(1+1k)=k+1.(Induction Hypothesis)\left(1+\frac11\right)\cdots\left(1+\frac1k\right)=k+1. \qquad \text{(Induction Hypothesis)}

We must show the product up to r=k+1r=k+1 equals k+2k+2. …

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