Help proving that $(n+a)^b = \Theta(n^b)$ Please you apologize me by my English.
I don't know how make that:
$$(n+a)^b = \Theta(n^b), b > 0$$
I know, I must to find two constants such that:
$$ c_{1} n^b \leq (n+a)^b \leq c_{2} n^b $$
I do not know what else to do. I'v tried with the Newton's binomial, but I'm lost.
 A: If $a \geq 0$ then 
$$(n+0)^b \leq (n+a)^b \,,$$
thus $c_1=1$ works.
Also, for all $n \geq a$ you have
$$(n+a)^b \leq (2n)^b =2^b n^b \,.$$
Now, fixing 
$$c_2 = \max \{ 2^b, \frac{(n+1)^b}{n^b},...,  \frac{(n+n-1)^b}{n^b} \}$$
you get the desired inequality.
For $a \leq 0$ you can get the inequlities the other way around, excepting that you'll have an issue if $a$ is a negative integer (what happens if $n=-a$?). 
A: Here, $(n+a)^{b} = n^{b}
 + a^{b}
 + ... \geq
 n^{b}
  \Longrightarrow
 (n+a)^{b}
 = \Omega
 (n^{b}
 ),$
$ Also, by  \space definition, f(n) = O (g(n)), n \rightarrow
 \infty
 \Longrightarrow
 |f(n)| \leq
  M(g(n)) $
$ As \space n \rightarrow
 \infty
 (n >> a), (n + a) \leq
  2n $
$(n+a)^{b}
 \leq(2n)^{b}\leq k*n^{b}\Longrightarrow
 (n+a)^{b}
 = O
 (n^{b}
 ), $
$Since (n+a)^{b}
 = \Omega
 (n^{b}
 ), and (n+a)^{b}
 = O
 (n^{b}
 ),$
$\Longrightarrow(n+a)^{b}
 = \Theta
 (n^{b}
 )$
A: For the upper bound,
$(n+a)^b = n^b(1+a/n)^b
$. 
If $n/a > b$,  $(1+a/n)^b < (1+1/b)^b < e$,
so we can take $c_2 = e$ for large enough $n$.
By taking $n$ large enough compared to $a$ and $b$,
we can take any value greater than 1 for $c_2$.
A: I proved it using the binomial expansion , which is
(x+y)n=nC0xny0+nC1xn−1y1+nC2xn−2y2+......nCnx0yn
now for the above problem (n+a)b=bC0nba0+nC1nb−1a1+nC2nb−2a2+......nCnn0an
we also now that for any polynomial
a0x1+a1x2+a2x2+⋯+anxn≤(a0+a1+a2+⋯+an)xn
substituting bC0a0
with C0, bC1a1 with C1
and so on (because that is a constant) we would get
(n+a)b=C0nb+C1nb−1+C2nb−2+......Cnn0
now i would jump to the following conclusion
C0nb≤C0nb+C1nb−1+C2nb−2+......Cnn0≤(C0+C1+C2+......Cn)nb
which makes (n+a)b=θ(nb)
