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418531 ภาคต้น 2552/โจทย์ปัญหาการพิสูจน์ II/เฉลยข้อ 4 - ประวัติรุ่นแก้ไข
2026-05-02T14:25:05Z
ประวัติรุ่นแก้ไขของหน้านี้ในวิกิ
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Cardcaptor: สร้างหน้าใหม่: == ข้อย่อย 1 == <math> a_n = \begin{cases} -2, & n = 0\\ a_{n-1} + 4, & n > 0 \end{cases} </math> == ข้อย่อย 2 == <math> a_n = ...
2009-07-09T20:15:47Z
<p>สร้างหน้าใหม่: == ข้อย่อย 1 == <math> a_n = \begin{cases} -2, & n = 0\\ a_{n-1} + 4, & n > 0 \end{cases} </math> == ข้อย่อย 2 == <math> a_n = ...</p>
<p><b>หน้าใหม่</b></p><div>== ข้อย่อย 1 ==<br />
<math><br />
a_n = <br />
\begin{cases}<br />
-2, & n = 0\\<br />
a_{n-1} + 4, & n > 0<br />
\end{cases}<br />
</math><br />
<br />
== ข้อย่อย 2 ==<br />
<math><br />
a_n = <br />
\begin{cases}<br />
0, & n = 0\\<br />
a_{n-1} + 2n, & n > 0<br />
\end{cases}<br />
</math><br />
<br />
== ข้อย่อย 3 ==<br />
<math><br />
a_n = <br />
\begin{cases}<br />
2, & n = 0\\<br />
a_{n-1} + 2 \cdot (-1)^n, & n > 0<br />
\end{cases}<br />
</math><br />
<br />
== ข้อย่อย 4 ==<br />
<math><br />
a_n = <br />
\begin{cases}<br />
0, & n = 0\\<br />
a_{n-1} + 2n - 1, & n > 0<br />
\end{cases}<br />
</math></div>
Cardcaptor