Update 文章 “使用归纳法证明二项式定理”

content/blog/使用归纳法证明二项式定理.md

@@ -22,14 +22,14 @@ $$
 试图使左右两边相等，需要使得左边分母为$k!(n+1-k)!$，所以通分得：
 
 $$
-\begin{aligned}
+\begin{align}
 & \frac{n!}{(k-1)!(n-k+1)!} + \frac{n!}{k!(n-k)!} \\\\
-&= \frac{n!k}{k!(n-k+1)!} + \frac{n!(n-k+1)}{k!(n-k+1)!} \\
+&= \frac{n!k}{k!(n-k+1)!} + \frac{n!(n-k+1)}{k!(n-k+1)!} \\\\
-&= \frac{n!k+n!(n-k+1)}{k!(n-k+1)!} \\
+&= \frac{n!k+n!(n-k+1)}{k!(n-k+1)!} \\\\
-&= \frac{n!(n+1)}{k!(n-k+1)!} \\
+&= \frac{n!(n+1)}{k!(n-k+1)!} \\\\
-&= \frac{(n+1)!}{k!(n+1-k)!} \\
+&= \frac{(n+1)!}{k!(n+1-k)!} \\\\
-&= \binom{n+1}{k}. \blacksquare
+&= \binom{n+1}{k}. \tag*{$\blacksquare$}
-\end{aligned}
+\end{align}
 $$
 
 随后证明二项式定理
@@ -44,13 +44,13 @@ $$(a+b)^0 = 1 = \sum_{k=0}^{0}\binom{0}{k}a^{0-k}b^k = a^0b^0.$$
 
 $$
 \begin{aligned}
-& (a+b)^{n+1} \\
+& (a+b)^{n+1} \\\\
-&= (a+b)(a+b)^n \\
+&= (a+b)(a+b)^n \\\\
-&= (a+b) \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k \\
+&= (a+b) \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k \\\\
-&= \sum_{k=0}^{n} \binom{n}{k} (a^{n+1-k}b^k + a^{n-k}b^{k+1}) \\
+&= \sum_{k=0}^{n} \binom{n}{k} (a^{n+1-k}b^k + a^{n-k}b^{k+1}) \\\\
-&= \sum_{k=0}^{n} \binom{n}{k} a^{n+1-k}b^k + \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^{k+1} \\
+&= \sum_{k=0}^{n} \binom{n}{k} a^{n+1-k}b^k + \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^{k+1} \\\\
-&= a^{n+1} + \sum_{k=1}^{n} \binom{n}{k} a^{n+1-k}b^k + b^{n+1} + \sum_{k=0}^{n-1} \binom{n}{k} a^{n-k}b^{k+1} \\
+&= a^{n+1} + \sum_{k=1}^{n} \binom{n}{k} a^{n+1-k}b^k + b^{n+1} + \sum_{k=0}^{n-1} \binom{n}{k} a^{n-k}b^{k+1} \\\\
-&= a^{n+1} + \sum_{k=1}^{n} \binom{n}{k} a^{n+1-k}b^k + b^{n+1} + \sum_{k=1}^{n} \binom{n}{k-1} a^{n-(k-1)}b^{(k-1)+1}. \\
+&= a^{n+1} + \sum_{k=1}^{n} \binom{n}{k} a^{n+1-k}b^k + b^{n+1} + \sum_{k=1}^{n} \binom{n}{k-1} a^{n-(k-1)}b^{(k-1)+1}. \\\\
 \end{aligned}
 $$
 
@@ -58,8 +58,8 @@ $$
 
 $$
 \begin{aligned}
-& a^{n+1} + \sum_{k=1}^{n} \binom{n}{k} a^{n+1-k}b^k + b^{n+1} + \sum_{k=1}^{n} \binom{n}{k-1} a^{n-(k-1)}b^{(k-1)+1}. \\
+& a^{n+1} + \sum_{k=1}^{n} \binom{n}{k} a^{n+1-k}b^k + b^{n+1} + \sum_{k=1}^{n} \binom{n}{k-1} a^{n-(k-1)}b^{(k-1)+1}. \\\\
-&= a^{n+1} + \sum_{k=1}^{n} \binom{n}{k} a^{n+1-k}b^k + b^{n+1} + \sum_{k=1}^{n} \binom{n}{k-1} a^{n+1-k}b^{k}. \\
+&= a^{n+1} + \sum_{k=1}^{n} \binom{n}{k} a^{n+1-k}b^k + b^{n+1} + \sum_{k=1}^{n} \binom{n}{k-1} a^{n+1-k}b^{k}. \\\\
 &= a^{n+1} + b^{n+1} + \sum_{k=1}^{n} \left[ \binom{n}{k} + \binom{n}{k-1} \right] a^{n+1-k}b^{k} .
 \end{aligned}
 $$
@@ -68,8 +68,8 @@ $$
 
 $$
 \begin{aligned}
-& a^{n+1} + b^{n+1} + \sum_{k=1}^{n} \left[ \binom{n}{k} + \binom{n}{k-1} \right] a^{n+1-k}b^{k} \\
+& a^{n+1} + b^{n+1} + \sum_{k=1}^{n} \left[ \binom{n}{k} + \binom{n}{k-1} \right] a^{n+1-k}b^{k} \\\\
-&= a^{n+1} + b^{n+1} + \sum_{k=1}^{n} \binom{n+1}{k} a^{n+1-k}b^{k} \\
+&= a^{n+1} + b^{n+1} + \sum_{k=1}^{n} \binom{n+1}{k} a^{n+1-k}b^{k} \\\\
 &= \sum_{k=0}^{n+1} \binom{n+1}{k} a^{n+1-k}b^{k}.
 \end{aligned}
 $$
@@ -77,9 +77,9 @@ $$
 所以，
 
 $$
-\begin{aligned}
+\begin{align}
-(a+b)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} a^{n+1-k}b^{k}. \blacksquare
+(a+b)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} a^{n+1-k}b^{k}. \tag*{$\blacksquare$}
-\end{aligned}
+\end{align}
 $$
 
 虽然在这里写得洋洋洒洒，但是说实话，在下天资愚钝，第一次见到这个证明，还是没那么容易理解的. 而且很有可能，现在证得出来，将来某天就会忘掉. 所以数学的学习是一个不断积累的过程，现在吃下的每一口，将来都会成长为构成身体的肌肉和骨骼.