Equivalent Expressions

Show that replacing k with k + 1 in 1 - 1/2^k gives an expression equivalent to 1 - 1/2^k + 1/2^(k+1)

1  -  1/ [2^(k + 1)]  =

1 - 1 / [2^k * 2] =

1 - 1/2^k (1/2)  =

1 + (-1/2)(1/2^k) =                     [notice that  -1/2  =  -1  + 1/2]

1 + [ -1 + 1/2] [ 1/2^k]  =

1 - 1/2^k  + (1/2)(1/2^k)  =

1 - 1/2^k + 1 / [ 2^k *2]  =

1 - 1/2^k +  1/ 2^(k + 1)

Hi Shades :)

Show that replacing k with k + 1 in 1 - 1/2k gives an expression equivalent to 1 - 1/2k + 1/2k+1

Going back to front it is easy enough.

$1-\frac{1}{2^k}+ \frac{1}{2 ^{k+1} } \\ =1-\left[\frac{1}{2^k}- \frac{1}{2 ^{k+1} } \right]\\ =1-\left[\frac{2}{2^{k+1}}- \frac{1}{2 ^{k+1} } \right]\\ =1-\left[\frac{1}{2^{k+1} } \right]\\ =1-\frac{1}{2^{k+1} } \\$

Now I want to work out how to do it in the other direction ://

Show that replacing k with k + 1 in 1 - 1/2k gives an expression equivalent to 1 - 1/2k + 1/2k+1

We can also do this with partial fractions and then it will go in the correct direction.

$1-\frac{1}{2^{k+1}}\\ =1+\;\;\frac{-1}{2*2^k}\\ \mbox{There are some integers A and B such that}\\ =1+\;\;\frac{A}{2^k}+\frac{B}{2^{k+1}}\\ =1+\;\;\frac{2A}{2^{k+1}}+\frac{B}{2^{k+1}}\\ =1+\;\;\frac{2A}{2^{k+1}}+\frac{B}{2^{k+1}}\\ =1+\;\;\frac{2A+B}{2^{k+1}}\\ \qquad SO\\ \qquad 2A+B=-1\\ \qquad \mbox{One solution to this is }\;\;A=-1\;\;and\;\;B=+1\\ =1+\;\;\frac{-1}{2^k}+\frac{1}{2^{k+1}}\\ =1-\;\;\frac{1}{2^k}+\frac{1}{2^{k+1}}\\ $

[Any A and B such that    2A+B= -1   will also be true ]

Yours is a really nice alternate solution Chris   

The Queen as returned!! * bows  *

Thanks Hayley,

I still do not expect to be as conscientious as before - I still have many other things cluttering up the "to do" list ://