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Exercise 4.4 · Q7

Q.Using the Mathematical induction, show that for any natural number nn,
[!FORMULA] 11.2.3+12.3.4+13.4.5+⋯+1n.(n+1).(n+2)=n(n+3)4(n+1)(n+2).\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\cdots+\dfrac{1}{n.(n+1).(n+2)} = \dfrac{n(n+3)}{4(n+1)(n+2)}.

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Let P(n):11.2.3+⋯+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)P(n):\dfrac1{1.2.3}+\cdots+\dfrac1{n(n+1)(n+2)}=\dfrac{n(n+3)}{4(n+1)(n+2)}.

Step 1. Base case. P(1)P(1): LHS =11.2.3=16=\dfrac1{1.2.3}=\dfrac16; RHS =1⋅44⋅2⋅3=424=16=\dfrac{1\cdot4}{4\cdot2\cdot3}=\dfrac{4}{24}=\dfrac16. True.

Step 2. Inductive hypothesis. Assume P(k): sum to kth term=k(k+3)4(k+1)(k+2)P(k):\ \text{sum to }k\text{th term}=\dfrac{k(k+3)}{4(k+1)(k+2)}. …

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