Q:

Finding the value of $1.1^{82}$ using $(1+x)^{82}$ to a certain accuracy

I found this question in a book. 
How many terms of the Maclaurin expansion of $(1+x)^{82}$ are needed to guarantee finding a value of $1.1^{82}$ to an accuracy of $10^{-6}$?
This is how I tried to do it:
We want the Lagrange error to be less than $10^{-6}$:
$$R_{n}(0.1)<10^{-6}$$
Now to find $R_{n}(0.1)$:
$$R_{n}(0.1)=\frac{f^{(n+1)}(0.1)\cdot0.1^{n+1}}{(n+1)!}$$
After writing the values of ${f^{(n+1)}}(x)$, I find a pattern and say that
$${f^{(n+1)}}(x)=\frac{82!}{(82-n)!}(1+x)^{82-n}$$
$${f^{(n+1)}}(0.1)=\frac{82!}{(82-n)!}(1.1)^{82-n}=\frac{82!(1.1)^{82-n}}{(82-n)!}$$
Now just substitute all of this back to $R_{n}(0.1)<10^{-6}$:
$$\frac{\frac{82!(1.1)^{82-n}}{(82-n)!}\cdot0.1^{n+1}}{(n+1)!}<10^{-6}$$
Great! Now I have an inequality that I can solve. I can't graph this so I just use trial and error (which seems like quite an intensive way to do it, so if anyone has a better way I'd love to know).
$$\frac{\frac{82!(1.1)^{82-n}}{(82-n)!}\cdot0.1^{n+1}}{(n+1)!}-10^{-6}<0$$
This seems to work first for $n=28$. That means the highest power of $x$ that I need is 28, so the number of terms in the expansion needs to be $n+1=29$. But the book's answer is 

 30

What am I doing wrong here? I'm still quite new to the power series expansions so if there's anything unclear in my working, please let me know (also, as a side note, the book itself may be wrong, though that isn't very likely I'd assume).

A:

Here is my expanded comment: Your formula for the derivatives is off-by-one; it should be
$${f^{(n)}}(x)=\frac{82!}{(82-n)!}(1+x)^{82-n}$$
and therefore
$$R_{n}(0.1)=\frac{f^{(n+1)}(0.1)\cdot 0.1^{n+1}}{(n+1)!}
=\frac{82!}{(82-n-1)!}(1.1)^{82-n-1}\frac{0.1^{n+1}}{(n+1)!}$$
With this you get $R_{29}=3.16\cdot 10^{-6}$ and $R_{30}=0.48\cdot 10^{-6}.$

