0
$\begingroup$

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.

2 Answers 2

1

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...

  • 0
    @MIH1406 using the distributive law: $( \color{darkgreen} b\cdot \color{maroon}a+ \color{darkgreen}c\cdot \color{maroon}a)=\color{maroon}a(\color{darkgreen}b+\color{darkgreen}c)$.2012-04-02
1

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