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I have: $(k+1)\bigr((k+1)!\bigl)+(k+1)!-1$

And want it to reach this form: $\bigr((k+1)+1\bigl)!-1$

Please, if possible go through the answer step by step.

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2 Answers 2

up vote 1 down vote accepted

First, factor out a $(k+1)!$ from the first two terms: $$\eqalign{ (k+1)\bigl((k+1)!\bigr)+(k+1)!-1&= \Bigl[\,\color{darkgreen}{(k+1) }\bigl(\color{maroon}{ (k+1)!}\bigr)+\color{darkgreen}1\cdot\color{maroon}{(k+1)!}\Bigr]-1\cr &= \color{maroon}{(k+1)!}\bigl[\, \color{darkgreen}{ (k+1)}+\color{darkgreen}1 \,\bigr]-1\cr &=(k+1)![ \, k+2\, ]-1\cr &=(k+2)!-1 \cr &=\ \cdots. } $$ I've left the last step for you...

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From where this (k+1)+1 came from in the first step? –  MIH1406 Apr 2 '12 at 17:22
    
@MIH1406 using the distributive law: $( \color{darkgreen} b\cdot \color{maroon}a+ \color{darkgreen}c\cdot \color{maroon}a)=\color{maroon}a(\color{darkgreen}b+\color{darkgreen}c)$. –  David Mitra Apr 2 '12 at 17:24

Hint: distribute the first $k+1$, then make it into $k+2-1$

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