Prove $\frac {1} {1+x^2} = \sum^{n-1}_{j=0} (-1)^j x^{2j} + (-1)^n \frac {x^{2n}} {1+x^2}, x\in \mathbb R, n \in \mathbb N$ by induction Prove $\frac {1} {1+x^2} = \sum^{n-1}_{j=0} (-1)^j x^{2j} + (-1)^n \frac {x^{2n}} {1+x^2}, x\in \mathbb R, n \in \mathbb N$ by induction.
$n = 1$: $\sum^{(1)-1}_{j=0} (-1)^j x^{2j} + (-1)^n \frac {x^{2n}} {1+x^2} = 1 -\frac {x^{2}} {1+x^2} = \frac {1} {1+x^2}$.
Now assume the identity is true for $n \le k$. 
We show the identity holds for $n=k+1$:
$\sum^{(k+1)-1}_{j=0} (-1)^j x^{2j} + (-1)^{k+1} \frac {x^{2(k+1)}} {1+x^2} = \sum^{k}_{j=0} (-1)^j x^{2j} + (-1)^kx^{2k} + (-1)^{k+1} \frac {x^{2(k+1)}} {1+x^2} = \frac {1} {1+x^2} + (-1)^kx^{2k} + (-1)^{k+1} \frac {x^{2(k+1)}} {1+x^2}$
where I've used the induction hypotheses in the last equality. However I cannot show that this sum equals $\frac {1} {1+x^2}$.
Thanks for your time.
 A: The equation in question is
$\dfrac{1}{1 + x^2} = \sum_{j = 0}^{n - 1} (-1)^j x^{2j} + (-1)^n \dfrac{x^{2n}}{1 + x^2} \tag{1}$
which, upon multiplication by $1 + x^2$ is seen to be equivalent to
$1 = (1 + x^2) \sum_{j = 0}^{n - 1} (-1)^j x^{2j} + (-1)^n x^{2n} \tag{2}$
or
$1 - (-1)^n x^{2n} = (1 + x^2) \sum_{j = 1}^{n - 1} (-1)^j x^{2j}; \tag{3}$
we prove (3) by induction on $n$.  It is easy to see that,  for $n = 1$, (3) reduces to the identity
$1 + x^2 =  1 + x^2, \tag{4}$
since the sum on the right of (3) clearly takes the value $1$ for $n = 1$.  Suppose then that (3) holds for some positive integer $k$:
$1 - (-1)^k x^{2k} = (1 + x^2) \sum_{j = 1}^{k - 1} (-1)^j x^{2j}; \tag{5}$
we subtract $(-1)^{k + 1}x^{2(k + 1)}$ from, and add $(-1)^k x^{2k}$ to, each side of (5), yielding
$1 - (-1)^{k + 1} x^{2(k + 1)} = (1 + x^2) \sum_{j = 1}^{k - 1} (-1)^j x^{2j} + (-1)^k x^{2k} - (-1)^{k + 1}x^{2(k + 1)} ; \tag{6}$
examining the last two terms on the right-hand side of (6), we see that
$(-1)^k x^{2k} - (-1)^{k + 1}x^{2(k + 1)} = (-1)^k x^{2k} (1 - (-1)x^2) = (-1)^k x^{2k} (1 + x^2), \tag{7}$
and thus we have
$(1 + x^2) \sum_{j = 1}^{k - 1} (-1)^j x^{2j} + (-1)^k x^{2k} - (-1)^{k + 1}x^{2(k + 1)}$
$= (1 + x^2) \sum_{j = 1}^{k - 1} (-1)^j x^{2j} + (-1)^k x^{2k} (1 + x^2) = (1 + x^2)\sum_{j = 1}^k (-1)^j x^{2j}, \tag{8}$
which shows that (6) becomes
$1 - (-1)^{k + 1} x^{2(k + 1)} = (1 + x^2)\sum_{j = 1}^k (-1)^j x^{2j}, \tag{9}$
completing the inductive proof of (3).  If we now add $(-1)^n x^{2n}$ to each side of (3), and then divide by $(1 + x^2)$, we obtain
$\dfrac{1}{1 + x^2} = \sum_{j = 1}^{n -1} (-1)^j x^{2j} + \dfrac{(-1)^n x^{2n}}{1 + x^2}, \tag{10}$
thus establising (1).  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
