I am now going to do some of the same alignments in AMS-TEX. Note the subtle differences in equation labeling.

Here is an example of the align environment.

(1)  (a + b)^{n+1} = (a + b)(a + b)^{n} = (a + b) ∑_{j=0}^{n} \binom{n}{j}a^{n−1}b^{j}

= ∑_{j=0}^{n} \binom{n}{j}a^{n+1−j}b^{j} + ∑_{j=1}^{n} \binom{n}{j − 1}a^{n−j}b^{j} (2)

= ∑_{j=0}^{n} \binom{n + 1}{j}a^{n+1−j}b^{j} (3)

(a + b)^{n+1} = (a + b)(a + b)^{n} = (a + b) ∑_{j=0}^{n} \binom{n}{j}a^{n−1}b^{j}  (1)

= ∑_{j=0}^{n} \binom{n}{j}a^{n+1−j}b^{j} + ∑_{j=1}^{n} \binom{n}{j − 1}a^{n−j}b^{j}  (2)

= ∑_{j=0}^{n} \binom{n + 1}{j}a^{n+1−j}b^{j}  (3)

Here is an example of the aligned environment.

(5–4)  \begin{aligned}(a + b)^{n+1} & = (a + b)(a + b)^{n} = (a + b) ∑_{j=0}^{n} \binom{n}{j}a^{n−1}b^{j} \\ & = ∑_{j=0}^{n} \binom{n}{j}a^{n+1−j}b^{j} + ∑_{j=1}^{n} \binom{n}{j − 1}a^{n−j}b^{j} \\ & = ∑_{j=0}^{n} \binom{n + 1}{j}a^{n+1−j}b^{j} \end{aligned}

\begin{aligned}(a + b)^{n+1} & = (a + b)(a + b)^{n} = (a + b) ∑_{j=0}^{n} \binom{n}{j}a^{n−1}b^{j} \\ & = ∑_{j=0}^{n} \binom{n}{j}a^{n+1−j}b^{j} + ∑_{j=1}^{n} \binom{n}{j − 1}a^{n−j}b^{j} \\ & = ∑_{j=0}^{n} \binom{n + 1}{j}a^{n+1−j}b^{j} \end{aligned}  (5–4)

Here is an example of the split environment.

\begin{aligned}(a + b)^{n+1} & = (a + b)(a + b)^{n} = (a + b) ∑_{j=0}^{n} \binom{n}{j}a^{n−1}b^{j} \\ & = ∑_{j=0}^{n} \binom{n}{j}a^{n+1−j}b^{j} + ∑_{j=1}^{n} \binom{n}{j − 1}a^{n−j}b^{j} \\ & = ∑_{j=0}^{n} \binom{n + 1}{j}a^{n+1−j}b^{j} \end{aligned} (1–2)

\begin{aligned}(a + b)^{n+1} & = (a + b)(a + b)^{n} = (a + b) ∑_{j=0}^{n} \binom{n}{j}a^{n−1}b^{j} \\ & = ∑_{j=0}^{n} \binom{n}{j}a^{n+1−j}b^{j} + ∑_{j=1}^{n} \binom{n}{j − 1}a^{n−j}b^{j} \end{aligned}

= ∑_{j=0}^{n} \binom{n + 1}{j}a^{n+1−j}b^{j}  (1–2)

Here is an example of the gather environment.

(a + b)^{n+1} = (a + b)(a + b)^{n} = (a + b) ∑_{j=0}^{n} \binom{n}{j}a^{n−1}b^{j} (1)

= ∑_{j=0}^{n} \binom{n}{j}a^{n+1−j}b^{j} + ∑_{j=1}^{n} \binom{n}{j − 1}a^{n−j}b^{j} (2)

= ∑_{j=0}^{n} \binom{n + 1}{j}a^{n+1−j}b^{j} (3)

(a + b)^{n+1} = (a + b)(a + b)^{n} = (a + b) ∑_{j=0}^{n} \binom{n}{j}a^{n−1}b^{j}  (1)

= ∑_{j=0}^{n} \binom{n}{j}a^{n+1−j}b^{j} + ∑_{j=1}^{n} \binom{n}{j − 1}a^{n−j}b^{j}  (2)

= ∑_{j=0}^{n} \binom{n + 1}{j}a^{n+1−j}b^{j}  (3)

Here is an example of the multline environment.

(1–2) (a + b)^{n+1} = (a + b)(a + b)^{n}

= (a + b) ∑_{j=0}^{n} \binom{n}{j}a^{n−1}b^{j} = ∑_{j=0}^{n} \binom{n}{j}a^{n+1−j}b^{j} + ∑_{j=1}^{n} \binom{n}{j − 1}a^{n−j}b^{j} = ∑_{j=0}^{n} \binom{n + 1}{j}a^{n+1−j}b^{j}

(a + b)^{n+1} = (a + b)(a + b)^{n}

= (a + b) ∑_{j=0}^{n} \binom{n}{j}a^{n−1}b^{j} = ∑_{j=0}^{n} \binom{n}{j}a^{n+1−j}b^{j} + ∑_{j=1}^{n} \binom{n}{j − 1}a^{n−j}b^{j} = ∑_{j=0}^{n} \binom{n + 1}{j}a^{n+1−j}b^{j} (1–2)