Prove by method of induction, for all $n \in N$: $\dfrac{1}{3.5} + \dfrac{1}{5.7} + \dfrac{1}{7.9} + \ldots \text{ to } n \text{ terms} = \dfrac{n}{3(2n+3)}$.

Prove by method of induction, for all n ∈ N: (1)/(3.5) + (1)/(5.7) + (1)/(7.9) + … to n terms = (n)/(3(2n+3)).

EXERCISE 4.1 · Q9

n \in N

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Let $P(n):\dfrac1{3.5}+\dfrac1{5.7}+\cdots\text{ to }n\text{ terms}=\dfrac{n}{3(2n+3)}$ , $k$ th term $\dfrac1{(2k+1)(2k+3)}$ . Base: $n=1$ : L.H.S. $=\dfrac1{15}$ , R.H.S. $=\dfrac{1}{3(5)}=\dfrac1{15}$ ; holds. Hypothesis: assume true for $k$ . Step: add $\dfrac1{(2k+3)(2k+5)}$ : $\dfrac{k}{3(2k+3)}+\dfrac1{(2k+3)(2k+5)}=\dfrac{k(2k+5)+3}{3(2k+3)(2k+5)}=\dfrac{2k^2+5k+3}{3(2k+3)(2k+5)}=\dfrac{(2k+3)(k+1)}{3(2k+3)(2k+5)}=\dfrac{k+1}{3(2k+5)} …

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Q.Prove by method of induction, for all $n \in N$ : $\dfrac{1}{3.5} + \dfrac{1}{5.7} + \dfrac{1}{7.9} + \ldots \text{ to } n \text{ terms} = \dfrac{n}{3(2n+3)}$ .

Concept understanding — Principle of Mathematical Induction

Q.Prove by method of induction, for all n∈Nn \in Nn∈N: 13.5+15.7+17.9+… to n terms=n3(2n+3)\dfrac{1}{3.5} + \dfrac{1}{5.7} + \dfrac{1}{7.9} + \ldots \text{ to } n \text{ terms} = \dfrac{n}{3(2n+3)}3.51​+5.71​+7.91​+… to n terms=3(2n+3)n​.

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Q.Prove by method of induction, for all $n \in N$ : $\dfrac{1}{3.5} + \dfrac{1}{5.7} + \dfrac{1}{7.9} + \ldots \text{ to } n \text{ terms} = \dfrac{n}{3(2n+3)}$ .

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