Sorry for the title it may be a very bad one. The whole question is as following:  How many terms of the Maclaurin expansion of $(1+x)^{82}$ are needed to guarantee finding a value of $1.1^{82}$ accurate to within $10^{-6}$?
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 $10^{-6}$:  $$R_{n}(0.1)&lt;10^{-6}$$  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 $R_{n}(0.1)&lt;10^{-6}$:  $$\frac{\frac{82!(1.1)^{82-n}}{(82-n)!}\cdot0.1^{n+1}}{(n+1)!}&lt;10^{-6}$$  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).  $$\frac{\frac{82!(1.1)^{82-n}}{(82-n)!}\cdot0.1^{n+1}}{(n+1)!}-10^{-6}&lt;0$$  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).
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