First, we decompose $\frac{4n^3 - n^2 - n + 1}{n^6 - n^5 + n^4 - n^3 + n^2 - n}$ into partial fractions.  We factor the denominator:
\begin{align*}
n^6 - n^5 + n^4 - n^3 + n^2 - n &= n(n^5 - n^4 + n^3 - n^2 + n - 1) \\
&= n(n^4 (n - 1) + n^2 (n - 1) + (n - 1)) \\
&= n(n - 1)(n^4 + n^2 + 1) \\
&= n(n - 1)[(n^4 + 2n^2 + 1) - n^2] \\
&= n(n - 1)[(n^2 + 1)^2 - n^2] \\
&= n(n - 1)(n^2 + n + 1)(n^2 - n + 1).
\end{align*}Then by partial fractions,
\[\frac{4n^3 - n^2 - n + 1}{n(n - 1)(n^2 + n + 1)(n^2 - n + 1)} = \frac{A}{n} + \frac{B}{n - 1} + \frac{Cn + D}{n^2 + n + 1} + \frac{En + F}{n^2 - n + 1}\]for some constants $A,$ $B,$ $C,$ $D,$ $E,$ and $F.$

Multiplying both sides by $n(n - 1)(n^2 + n + 1)(n^2 - n + 1),$ we get
\begin{align*}
4n^3 - n^2 - n + 1 &= A(n - 1)(n^2 + n + 1)(n^2 - n + 1) \\
&\quad + Bn(n^2 + n + 1)(n^2 - n + 1) \\
&\quad + (Cn + D)n(n - 1)(n^2 - n + 1) \\
&\quad + (En + F)n(n - 1)(n^2 + n + 1).
\end{align*}Setting $n = 0,$ we get $-A = 1,$ so $A = -1.$

Setting $n = 1,$ we get $3B = 3,$ so $B = 1.$  The equation above then becomes
\begin{align*}
4n^3 - n^2 - n + 1 &= -(n - 1)(n^2 + n + 1)(n^2 - n + 1) \\
&\quad + n(n^2 + n + 1)(n^2 - n + 1) \\
&\quad + (Cn + D)n(n - 1)(n^2 - n + 1) \\
&\quad + (En + F)n(n - 1)(n^2 + n + 1).
\end{align*}This simplifies to
\[n^4 + 4n^3 - 2n^2 - n = (Cn + D)n(n - 1)(n^2 - n + 1) + (En + F)n(n - 1)(n^2 + n + 1).\]Dividing both sides by $n(n - 1),$ we get
\[-n^2 + 3n + 1 = (Cn + D)(n^2 - n + 1) + (En + F)(n^2 + n + 1).\]Expanding, we get
\[-n^2 + 3n + 1 = (C + E)n^3 + (C + D + E + F)n^2 + (C - D + E + F)n + D + F.\]Matching coefficients, we get
\begin{align*}
C + E &= 0, \\
-C + D + E + F &= -1, \\
C - D + E + F &= 3, \\
D + F &= 1.
\end{align*}Since $C + E = 0,$ $-D + F = 3.$  Hence, $D = -1$ and $F = 2.$  Then $-C + E = -2,$ so $C = 1$ and $E = -1.$  Therefore,
\[\frac{4n^3 - n^2 - n + 1}{n^6 - n^5 + n^4 - n^3 + n^2 - n} = \frac{1}{n - 1} - \frac{1}{n} + \frac{n - 1}{n^2 + n + 1} - \frac{n - 2}{n^2 - n + 1}.\]Then
\begin{align*}
\sum_{n = 2}^\infty \frac{4n^3 - n^2 - n + 1}{n^6 - n^5 + n^4 - n^3 + n^2 - n} &= \left( 1 - \frac{1}{2} + \frac{1}{7} \right) \\
&\quad + \left( \frac{1}{2} - \frac{1}{3} + \frac{2}{13} - \frac{1}{7} \right) \\
&\quad + \left( \frac{1}{3} - \frac{1}{4} + \frac{3}{21} - \frac{2}{13} \right) + \dotsb \\
&= 1.
\end{align*}
먼저 $\frac{4n^3 - n^2 - n + 1}{n^6 - n^5 + n^4 - n^3 + n^2 - n}$을 부분 분수로 분해합니다. 분모를 인수분해합니다:
\begin{align*}
n^6 - n^5 + n^4 - n^3 + n^2 - n &= n(n^5 - n^4 + n^3 - n^2 + n - 1) \}
&= n(n^4 (n - 1) + n^2 (n - 1) + (n - 1))
&= n(n - 1)(n^4 + n^2 + 1)
&= n(n - 1)[(n^4 + 2n^2 + 1) - n^2]
&= n(n - 1)[(n^2 + 1)^2 - n^2]
&= n(n - 1)(n^2 + n + 1)(n^2 - n + 1).
\end{align*}그런 다음 부분 분수로,
\[\frac{4n^3 - n^2 - n + 1}{n(n - 1)(n^2 + n + 1)(n^2 - n + 1)}{n(n - 1)(n^2 + n + 1)(n^2 - n + 1)} 일부 상수 $A,$ $B,$ $ $C,$ $ $D,$ $ $E,$ 및 $F.$에 대해 \[\frac{A}{n} + \frac{B}{n - 1} + \frac{Cn + D}{n^2 + n + 1} + \frac{En + F}{n^2 - n + 1}]= ${4n^3 - n + 1}{n(n^2 - n + 1)}{n(n^2 - n + 1)}($$$$$$$$$$$$$$$$)}입니다.

양변에 $n(n - 1)(n^2 + n + 1)(n^2 - n + 1)$을 곱하면 다음과 같이 구할 수 있습니다.
\begin{align*}
4n^3 - n^2 - n + 1 &= A(n - 1)(n^2 + n + 1)(n^2 - n + 1) \$.
&\쿼드 + Bn(n^2 + n + 1)(n^2 - n + 1) \$.
&\quad + (Cn + D)n(n - 1)(n^2 - n + 1) \$ &\quad + (Cn + D)(n - 1)
&\quad + (En + F)n(n - 1)(n^2 + n + 1).
\end{align*}$n = 0,$을 설정하면 $-A = 1,$이므로 $A = -1.$이 됩니다.

n = 1,$ 을 설정하면 $3B = 3,$ 이므로 $B = 1.$ 위의 방정식은 다음과 같이 됩니다.
\begin{align*}
4n^3 - n^2 - n + 1 &= -(n - 1)(n^2 + n + 1)(n^2 - n + 1) \.
&\쿼드 + n(n^2 + n + 1)(n^2 - n + 1) \.
&\quad + (Cn + D)n(n - 1)(n^2 - n + 1) \\.
&\quad + (En + F)n(n - 1)(n^2 + n + 1).
\end{align*}이것은 다음과 같이 단순화됩니다.
\[n^4 + 4n^3 - 2n^2 - n = (Cn + D)n(n - 1)(n^2 - n + 1) + (En + F)n(n - 1)(n^2 + n + 1).\]양변을 $n(n - 1)$로 나누면 다음과 같이 됩니다.
\[-n^2 + 3n + 1 = (Cn + D)(n^2 - n + 1) + (En + F)(n^2 + n + 1).\]를 확장하면 다음과 같습니다.
\[-n^2 + 3n + 1 = (C + E)n^3 + (C + D + E + F)n^2 + (C - D + E + F)n + D + F.\]일치하는 계수를 구하면 다음과 같습니다.
\begin{align*}
C + E &= 0, \
-C + D + E + F &= -1, \\.
C - D + E + F &= 3, \\.
D + F &= 1.
\end{align*}$C + E = 0$이므로 $D + F = 3.$ 따라서 $D = -1$이고 $F = 2.$ 그러면 $C + E = -2$이므로 $C = 1$이고 $E = -1.$ 입니다,
\frac{4n^3 - n^2 - n + 1}{n^6 - n^5 + n^4 - n^3 + n^2 - n} = \frac{1}{n - 1} - \frac{1}{n} + \frac{n - 1}{n^2 + n + 1} - \frac{2 - n + 2}{n^2 - n + 1}.\]그러면 다음과 같습니다.
\begin{align*}
\sum_{n = 2}^\infty \frac{4n^3 - n^2 - n + 1}{n^6 - n^5 + n^4 - n^3 + n^2 - n} &= \left( 1 - \frac{1}{2} + \frac{1}{7} \right) \.
&\quad + \left( \frac{1}{2} - \frac{1}{3} + \frac{2}{13} - \frac{1}{7} \right) \\.
&\quad + \left( \frac{1}{3} - \frac{1}{4} + \frac{3}{21} - \frac{2}{13} \right) + \dotsb \.
&= 1.
\end{align*}