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Exercise 4.4 · Q8

Q.Using the Mathematical induction, show that for any natural number nn,
[!FORMULA] 12.5+15.8+18.11+⋯+1(3n−1)(3n+2)=n6n+4.\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\cdots+\dfrac{1}{(3n-1)(3n+2)} = \dfrac{n}{6n+4}.

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Let P(n):12.5+15.8+⋯+1(3n−1)(3n+2)=n6n+4P(n):\dfrac1{2.5}+\dfrac1{5.8}+\cdots+\dfrac1{(3n-1)(3n+2)}=\dfrac n{6n+4}.

Step 1. Base case. P(1)P(1): LHS =12⋅5=110=\dfrac1{2\cdot5}=\dfrac1{10}; RHS =110=\dfrac1{10}. True.

Step 2. Inductive hypothesis. Assume P(k): sum to kth term=k6k+4P(k):\ \text{sum to }k\text{th term}=\dfrac k{6k+4}. …

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