We just learned induction, and I'm already stuck. I'll try my best to put my work into text.
1 − 2 + 2^2 − 2^3 + ... + (−1)^n(2^n) = (2^(n+1)(−1)^n + 1)/3
for all positive integers n.
Already have basis-step complete.
Inductive Step: Assume n=k, then...
1 − 2 + 2^2 − 2^3 + ... + (−1)^k(2^k) = (2^(k+1)(−1)^k + 1)/3
then k+1
1 − 2 + 2^2 − 2^3 + ... + (−1)^(k+1)(2^(k+1)) = (2^((k+1)+1)(−1)^(k+1) + 1)/3
(2^(k+1)(−1)^k + 1)/3 + (-1)^(k+1)2(k+1) = (2^(k+2)(−1)^(k+1) + 1)/3
(2^(k+1)(−1)^k + 1)/3 + ((-3)^(k+1)2(k+1))/3 = (2^(k+2)(−1)^(k+1) + 1)/3
This is where I start having trouble, re-arranging and dividing out things like 2^(k+1) or (-1)^(k+1) have led to nothing... I think.
I don't know if the work to the point is correct or not, but I've been mulling over this for quite a while now to no avail.
First, show that this is true for $n=1$:
$\sum\limits_{k=0}^{1}(-1)^{k}(2^{k})=\frac{(−1)^{1}(2^{1+1})+1}{3}$
Second, assume that this is true for $n$:
$\sum\limits_{k=0}^{n}(-1)^{k}(2^{k})=\frac{(−1)^{n}(2^{n+1})+1}{3}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=0}^{n+1}(-1)^{k}(2^{k})=$
$\color\red{\sum\limits_{k=0}^{n}(-1)^{k}(2^{k})}+(-1)^{n+1}(2^{n+1})=$
$\color\red{\frac{(−1)^{n}(2^{n+1})+1}{3}}+(-1)^{n+1}(2^{n+1})=$
$\frac{(−1)^{n}(2^{n+1})+1}{3}+\frac{3(-1)^{n+1}(2^{n+1})}{3}=$
$\frac{(−1)^{n}(2^{n+1})+1+3(-1)^{n+1}(2^{n+1})}{3}=$
$\frac{(−1)^{n}(2^{n+1})+3(-1)^{n+1}(2^{n+1})+1}{3}=$
$\frac{(−1)^{n}(2^{n+1})(1+3(-1)^{1})+1}{3}=$
$\frac{(−1)^{n}(2^{n+1})(1-3)+1}{3}=$
$\frac{(−1)^{n}(2^{n+1})(-2)+1}{3}=$
$\frac{(−1)^{n}(2^{n+1})(-1)(2)+1}{3}=$
$\frac{(−1)^{n}(-1)(2^{n+1})(2)+1}{3}=$
$\frac{(−1)^{n+1}(2^{n+1+1})+1}{3}$
Please note that the assumption is used only in the part marked red.