Prove by method of induction, for all $n \in N$: $\dfrac{1}{1.3} + \dfrac{1}{3.5} + \dfrac{1}{5.7} + \ldots + \dfrac{1}{(2n-1)(2n+1)} = \dfrac{n}{2n+1}$.

Prove by method of induction, for all n ∈ N: (1)/(1.3) + (1)/(3.5) + (1)/(5.7) + … + (1)/((2n-1)(2n+1)) = (n)/

EXERCISE 4.1 · Q8

n \in N

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Let $P(n):\dfrac{1}{1.3}+\dfrac{1}{3.5}+\cdots+\dfrac{1}{(2n-1)(2n+1)}=\dfrac{n}{2n+1}$ . Base: $n=1$ : L.H.S. $=\dfrac13$ , R.H.S. $=\dfrac13$ ; holds. Hypothesis: assume true for $k$ . Step: add $\dfrac1{(2k+1)(2k+3)}$ : $\dfrac{k}{2k+1}+\dfrac1{(2k+1)(2k+3)}=\dfrac{k(2k+3)+1}{(2k+1)(2k+3)}=\dfrac{2k^2+3k+1}{(2k+1)(2k+3)}=\dfrac{(2k+1)(k+1)}{(2k+1)(2k+3)}=\dfrac{k+1}{2k+3}=\dfrac{k+1}{2(k+1)+ …

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Q.Prove by method of induction, for all $n \in N$ : $\dfrac{1}{1.3} + \dfrac{1}{3.5} + \dfrac{1}{5.7} + \ldots + \dfrac{1}{(2n-1)(2n+1)} = \dfrac{n}{2n+1}$ .

Concept understanding — Principle of Mathematical Induction

Q.Prove by method of induction, for all n∈Nn \in Nn∈N: 11.3+13.5+15.7+…+1(2n−1)(2n+1)=n2n+1\dfrac{1}{1.3} + \dfrac{1}{3.5} + \dfrac{1}{5.7} + \ldots + \dfrac{1}{(2n-1)(2n+1)} = \dfrac{n}{2n+1}1.31​+3.51​+5.71​+…+(2n−1)(2n+1)1​=2n+1n​.

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Q.Prove by method of induction, for all $n \in N$ : $\dfrac{1}{1.3} + \dfrac{1}{3.5} + \dfrac{1}{5.7} + \ldots + \dfrac{1}{(2n-1)(2n+1)} = \dfrac{n}{2n+1}$ .

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