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Exercises · Q13

Q.Prove the following by using the principle of mathematical induction for all n∈Nn \in N: (1+31)(1+54)(1+79)…(1+(2n+1)n2)=(n+1)2\left(1+\dfrac{3}{1}\right)\left(1+\dfrac{5}{4}\right)\left(1+\dfrac{7}{9}\right)\ldots\left(1+\dfrac{(2n+1)}{n^2}\right) = (n+1)^2

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For a positive integer rr, the general factor is

1+2r+1r2=r2+2r+1r2=(r+1)2r2.1+\frac{2r+1}{r^2}=\frac{r^2+2r+1}{r^2}=\frac{(r+1)^2}{r^2}.

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

(1+31)(1+54)(1+79)⋯(1+2n+1n2)=(n+1)2.\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)\cdots\left(1+\frac{2n+1}{n^2}\right)=(n+1)^2.

Base case: For n=1n=1,

LHS=1+31=4,RHS=(1+1)2=4.\text{LHS}=1+\frac31=4,\qquad \text{RHS}=(1+1)^2=4.

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

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

(1+31)⋯(1+2k+1k2)=(k+1)2.(Induction Hypothesis)\left(1+\frac31\right)\cdots\left(1+\frac{2k+1}{k^2}\right)=(k+1)^2. \qquad \text{(Induction Hypothesis)}

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

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