# Mathematical Induction Arithmetic

• February 9th 2010, 05:36 PM
buddyp450
Mathematical Induction Arithmetic
I was working on proving using mathematical induction the premise:

P(n): 2 - 2*7 + 2*7^2 - ... + 2(-7)^n = (1-(-7)^(n+1))/4 while n>=0

I got to the point of proving P(n+1),

[1-(-7)^(n+1) + 8(-7)^(n+1)]/4

but I'm not sure arithmetically how that will simplify into the expected outcome of:

[1-(-7)^(n+2)]/4

???

Am I even right that it will simplify to that seeing as how that is what I was expecting out of the original P(n).
• February 9th 2010, 06:08 PM
Quote:

Originally Posted by buddyp450
I was working on proving using mathematical induction the premise:

P(n): 2 - 2*7 + 2*7^2 - ... + 2(-7)^n = (1-(-7)^(n+1))/4 while n>=0

I got to the point of proving P(n+1),

[1-(-7)^(n+1) + 8(-7)^(n+1)]/4

but I'm not sure arithmetically how that will simplify into the expected outcome of:

[1-(-7)^(n+2)]/4

???

Am I even right that it will simplify to that seeing as how that is what I was expecting out of the original P(n).

$2-2(7)+2(7^2)+..........+2(-7)^n=\frac{1-(-7)^{n+1}}{4}$ ??

If so, then

$2-2(7)+........+2(-7)^n+2(-7)^{n+1}\ must\ =\frac{1-(-7)^{n+2}}{4}$

Check:

$\frac{1-(-7)^{n+2}}{4}=\frac{1-(-7)^{n+1}(-7)}{4}=\frac{1-(-7)^{n+1}}{4}+\frac{8(-7)^{n+1}}{4}$ $=\frac{1-(-7)^{n+1}}{4}+2(-7)^{n+1}$

Hence, if true for some n, then definately true for n+1 because of that,
so the dominoes are all stacked up.

True for n=0 ?

$2=\frac{1-(-7)^1}{4}$ ?

Yes...... Proven
• February 10th 2010, 07:52 AM
novice
Quote:

Originally Posted by buddyp450
I was working on proving using mathematical induction the premise:

P(n): 2 - 2*7 + 2*7^2 - ... + 2(-7)^n = (1-(-7)^(n+1))/4 while n>=0

I got to the point of proving P(n+1),

[1-(-7)^(n+1) + 8(-7)^(n+1)]/4 <--this is correct

but I'm not sure arithmetically how that will simplify into the expected outcome of:

[1-(-7)^(n+2)]/4

???

Am I even right that it will simplify to that seeing as how that is what I was expecting out of the original P(n).

Working with the expression $
\frac{1}{4}\{1-(-7)^{n+1}+8(-7)^{n+1)}\}
$
, the tricky part here is the application of Distributive Law to a negative term. Now the positive integer 8 becomes negative.

$
\frac{1}{4}[1-(-7)^{n+1}(1-8)]
$
, which is $
\frac{1-(-7)^{n+2}}{4}
$
,