Verify by long division that $\frac {1}{1+t}= 1-t+t^2-t^3+....+(-1)^n t^n+\frac{(-1)^{n+1}t^{n+1}}{1+t}$ I'm asked to verify that $\dfrac {1}{1+t} = 1-t+t^2-t^3+....+(-1)^n t^n+\dfrac{(-1)^{n+1}t^{n+1}}{1+t}$
I've tried starting with term  $\dfrac{(-1)^{n+1}t^{n+1}}{1+t}$
and applying the long division, but I still didn't get the desired result.
 A: So you have to prove $1=(1+t)\Big( 1-t+t^2-t^3+....+(-1)^n t^n\Big)+(-1)^{n+1}t^{n+1}$
or $$1-(-1)^{n+1}t^{n+1} =(1+t)\Big( 1-t+t^2-t^3+....+(-1)^n t^n\Big)$$
or $${1-(-1)^{n+1}t^{n+1} \over 1+t} = 1-t+t^2-t^3+....+(-1)^n t^n$$
And this can be done with the sum of geometric progression. Do you know the formula?
A: By long division:
$\dfrac1{1+t}=1-\dfrac t{1+t}=1-t+\dfrac {t^2}{1+t}=1-t+t^2-\dfrac{t^3}{1+t}...$
A: Use
$\dfrac{1-t^n}{1-t}
=\sum_{k=0}^{n-1} t^k
$
(easily proved by induction
or cross-multiplying)
and put
$-t$ for $t$.
A: Alternatively, use induction. 
For $n=1$:
$$\frac{1}{1+t}=1-t+\frac{t^2}{1+t}=\frac{1-t^2+t^2}{1+t} \color{green}{\checkmark} $$
Assume it is true for $n=k$. 
$$1-t+t^2-t^3+....+(-1)^k t^k+\dfrac{(-1)^{k+1}t^{k+1}}{1+t}=\frac1{1+t}$$
Prove it for $n=k+1$:
$$1-t+t^2-t^3+....+(-1)^k t^k+(-1)^{k+1}t^{k+1}+\dfrac{(-1)^{k+2}t^{k+2}}{1+t}=\\
1-t+t^2-t^3+....+(-1)^k t^k+\dfrac{(-1)^{k+1}t^{k+1}+\require{cancel}\cancel{(-1)^{k+1}t^{k+2}}+\cancel{(-1)^{k+2}t^{k+2}}}{1+t}=\\
1-t+t^2-t^3+....+(-1)^k t^k+\dfrac{(-1)^{k+1}t^{k+1}}{1+t}=\frac1{1+t}.$$
