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Exercise 4.4 · Q6

Q.Using the Mathematical induction, show that for any natural number n≥2n\ge 2,
[!FORMULA] 11+2+11+2+3+11+2+3+4+⋯+11+2+3+⋯+n=n−1n+1.\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+\cdots+\dfrac{1}{1+2+3+\cdots+n} = \dfrac{n-1}{n+1}.

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Let P(n):11+2+11+2+3+⋯+11+2+⋯+n=n−1n+1P(n):\dfrac1{1+2}+\dfrac1{1+2+3}+\cdots+\dfrac1{1+2+\cdots+n}=\dfrac{n-1}{n+1}, n≥2n\ge2. Note 1+2+⋯+k=k(k+1)21+2+\cdots+k=\dfrac{k(k+1)}2, so each term is 2k(k+1)\dfrac2{k(k+1)}.

Step 1. Base case. P(2)P(2): LHS =11+2=13=\dfrac1{1+2}=\dfrac13; RHS =2−12+1=13=\dfrac{2-1}{2+1}=\dfrac13. True.

Step 2. Inductive hypothesis. Assume P(k)P(k) holds, with sum =k−1k+1=\dfrac{k-1}{k+1} up to the kkth term. …

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