tính tổng: $S=1+8q+27q^2+...+n^3q^{n-1}+...,(\left | q \right |< 1)$
#2
Posted 12-03-2013 - 00:28
Lời giải:
Xét hàm sinh ứng với $S$: \[
f\left( q \right) = \sum\limits_{k = 1}^\infty {k^3 q^{k - 1} }
\]
Đặt dãy $(x_n)$ với $x_n=n^3$ thì $f\left( q \right) = \sum\limits_{k = 1}^\infty {x_k q^{k - 1} }$. Ta sẽ tìm hệ thức truy hồi của $(x_n)$.
\[
\begin{array}{l}
x_n = n^3 \\
\left. \begin{array}{l}
x_{n + 1} = \left( {n + 1} \right)^3 = n^3 + 3n^2 + 3n + 1 \\
x_{n - 1} = \left( {n - 1} \right)^3 = n^3 - 3n^2 + 3n - 1 \\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}
x_{n + 1} + x_{n - 1} = 2x_{n - 1} + 6n \\
x_{n + 1} - x_{n - 1} = 6n^2 + 2 \\
\end{array} \right. \\
x_{n - 2} = \left( {n - 2} \right)^3 = n^3 - 6n^2 + 12n - 8 = x_n - \left( {x_{n + 1} - x_{n - 1} - 2} \right) + 2\left( {x_{n + 1} + x_{n - 1} - 2x_n } \right) - 8 \\
\Leftrightarrow x_{n - 2} = x_{n + 1} + 3x_{n - 1} - 3x_n - 6 \Leftrightarrow x_{n + 1} = 3x_n - 3x_{n - 1} + x_{n - 2} + 6 \\
\Leftrightarrow x_n = 3x_{n - 1} - 3x_{n - 2} + x_{n - 3} + 6 \quad \forall n \ge 4 \\
\end{array}
\]
Do đó, với mọi $q$ mà $|q|<1$:
\[
\begin{array}{rcl}
f\left( q \right) &=& \sum\limits_{k = 1}^\infty {k^3 q^{k - 1} } \\
&=& x_1 + x_2 q + x_3 q^2 + \sum\limits_{k = 4}^\infty {\left( {3x_{k - 1} - 3x_{k - 2} + x_{k - 3} + 6} \right)q^{k - 1} } \\
&=& 1 + 8q + 27q^2 + 3q\sum\limits_{k = 4}^\infty {x_{k - 1} q^{k - 2} } - 3q^2 \sum\limits_{k = 4}^\infty {x_{k - 2} q^{k - 3} } \\
&+& q^3 \sum\limits_{k = 4}^\infty {x_{k - 3} q^{k - 4} } + 6\left( {\sum\limits_{k = 1}^\infty {q^{k - 1} } - q^2 - q - 1} \right) \\
&=& 1 + 8q + 27q^2 + 3q\left[ {f\left( q \right) - x_1 - x_2 q} \right] - 3q^2 \left[ {f\left( q \right) - x_1 } \right] \\
&+& q^3 f\left( q \right) + \frac{6}{{1 - q}} - 6q^2 - 6q - 6 \\
\Rightarrow f\left( q \right)\left( {1 - 3q + 3q^2 + q^3 } \right) &=& - 5 - q + \frac{6}{{1 - q}} = \frac{{q^2 + 4q + 1}}{{1 - q}} \\
\Rightarrow f\left( q \right) &=& \frac{{q^2 + 4q + 1}}{{\left( {1 - q} \right)^4 }} \\
\end{array}
\]
Edited by perfectstrong, 12-03-2013 - 22:27.
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#3
Posted 19-03-2013 - 13:50
tính tổng: $S=1+8q+27q^2+...+n^3q^{n-1}+...,(\left | q \right |< 1)$
Xét tổng riêng
$S_n=\sum_{k=1}^n k^3q^{k-1}$
Ta có: $\Delta(q^{k-1})=(q-1)q^{k-1}$
$\Delta(k^3)=3k^2+3k+1$
$\Delta(3k^2+3k+1)=6(k+1)$
$\Delta(k+1)=1$
Do đó:
$(q-1)S_n=\sum_{k=1}^n k^3\Delta(q^{k-1})=k^3q^{k-1}\bigg|_{k=1}^{n+1}-\sum_{k=1}^n (3k^2+3k+1)q^k$
$(q-1)S_n=(n+1)^3q^n-1-\dfrac{1}{q-1}\sum_{k=1}^n(3k^2+3k+1)\Delta(q^k)$
$(q-1)S_n=(n+1)^3q^n-1-\dfrac{1}{q-1}(3k^2+3k+1)q^k\bigg|_{k=1}^{n+1}+\dfrac{6}{q-1}\sum_{k=1}^n(k+1)q^{k+1}$
$(q-1)S_n=(n+1)^3q^n+\dfrac{6q+1}{q-1}-\dfrac{(3n^2+9n+7)q^{n+1}}{q-1}$
$\qquad\qquad\qquad+\dfrac{6}{(q-1)^2}\sum_{k=1}^n(k+1)\Delta(q^{k+1})$
$(q-1)S_n=(n+1)^3q^n+\dfrac{6q+1}{q-1}-\dfrac{(3n^2+9n+7)q^{n+1}}{q-1}$
$\qquad\qquad\qquad+\dfrac{6}{(q-1)^2}(k+1)q^{k+1}\bigg|_{k=1}^{n+1}-\dfrac{6}{(q-1)^2}\sum_{k=1}^nq^{k+2}$
$(q-1)S_n=(n+1)^3q^n+\dfrac{6q+1}{q-1}-\dfrac{(3n^2+9n+7)q^{n+1}}{q-1}$
$\qquad\qquad\qquad+\dfrac{6(n+2)q^{n+2}}{(q-1)^2}-\dfrac{12q^2}{(q-1)^2}-\dfrac{6q^{n+3}}{(q-1)^3}+\dfrac{6q^3}{(q-1)^3}$
Từ đó ta có:
$S_n=\dfrac{q^2+4q+1}{(q-1)^4}+\dfrac{(n+1)^3q^n}{q-1}-\dfrac{(3n^2+9n+7)q^{n+1}}{(q-1)^2}+\dfrac{6(n+2)q^{n+2}}{(q-1)^3}$
Do $|q|<1$ nên $p=\left|\dfrac{1}{q}\right|>1$ và hàm mũ tiến tới vô cùng "nhanh" hơn hàm đa thức nên chuyển qua giới hạn ta có:
$\lim_{n\to +\infty} S_n=\dfrac{q^2+4q+1}{(q-1)^4}$
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#4
Posted 26-04-2013 - 19:53
tính tổng: $S=1+8q+27q^2+...+n^3q^{n-1}+...,(\left | q \right |< 1)$
Edited by nthoangcute, 26-04-2013 - 19:55.
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