Pascal's Triangle and the Binomial Theorem

Row 0 is just $1$. Every later row starts and ends with $1$, and every other entry is the sum of the two entries diagonally above it. Label the entries with $t_{n,r}$, meaning row $n$, position $r$, where BOTH are counted from 0. The rule is then a recursive formula: $t_{n,r}=t_{n-1,r-1}+t_{n-1,r}\text{.}$ For example, $t_{5,2}=t_{4,1}+t_{4,2}=4+6=10$. Counting from 0 matters. Row 5 is $1,5,10,10,5,1$ (six entries), so "row $n$" always has $n+1$ entries, and the very top 1 is row 0, not row 1.

• Symmetry: each row reads the same forwards and backwards.
• Row sum: the entries in row $n$ add up to $2^n$ (e.g. row 4: $1+4+6+4+1=16=2^4$).
• The second entry (and second-to-last) of row $n$ is always $n$ itself.
• The diagonal just inside the outer '1's gives the natural numbers $1,2,3,4,\dots$, and the next diagonal in gives the triangular numbers $1,3,6,10,15,\dots$
• Hockey stick: add entries down a diagonal and the total sits just below and to the side of the last one, e.g. $1+2+3+4=10$ (row 5).
• Powers of 11: rows 0 to 4 read as numbers give $1,11,121,1331,14641$, which are $11^0$ to $11^4$.

These patterns are not coincidences, and each one follows from the addition rule. Every entry in row $n$ is used twice when building row $n+1$ (once for the entry below-left, once for below-right), so the row total doubles every time: $1,2,4,8,\cdots =2^n$. Symmetry follows because the rule treats left and right identically and both edges start as 1. The second diagonal is $1,2,3,4,\dots$ because each entry is the one above plus 1 from the edge. The triangular numbers appear because each one is the running total of that diagonal, which is also why the hockey-stick pattern works. The powers-of-11 pattern is the binomial expansion in disguise: $11^n={(10+1)}^n$, so row $n$ gives the digits until an entry reaches 10 and has to carry (from row 5 on).

Why should a triangle of additions give the coefficients of a product? Look at how ${(a+b)}^3$ is built from ${(a+b)}^2=a^2+2ab+b^2$. Multiplying by $(a+b)$ means multiplying every term by $a$, multiplying every term by $b$, and collecting like terms. Each new coefficient receives exactly two contributions, one from each neighbouring coefficient of the row above. That is precisely Pascal's addition rule, so the coefficients of ${(a+b)}^n$ must be row $n$.

To expand ${(a+b)}^n$, use row $n$ of Pascal's triangle as the coefficients. The power of $a$ starts at $n$ and counts DOWN to 0; the power of $b$ starts at $0$ and counts UP to $n$ — so every term's exponents always add up to $n$, and there are $n+1$ terms in total. When $a$ or $b$ has a coefficient, as in ${(2x+3)}^4$, put the WHOLE binomial term in brackets before raising it to a power: ${(2x)}^3=8x^3$, not $2x^3$.

Pascal's triangle for binomial expansion | Algebra II | Khan Academy

Tests often ask for a single term or coefficient, such as the coefficient of $x^3$ in ${(2x+3)}^5$. You don't need the whole expansion. The term with $b^r$ is $t_{n,r}{a}^{n-r}{b}^r$, which is entry $r+1$ of row $n$ when counting from 1. Here $x^3$ comes from ${(2x)}^3$, so the second term appears to the power $5-3=2$ and $r=2$: the coefficient is $t_{5,2}{(2)}^3{(3)}^2=10(8)(9)=720\text{.}$ Always include the coefficient's power and the sign: in ${(x-2)}^5$, the $x^2$ term is $10x^2{(-2)}^3=-80x^2$. Which approach should you use? If the question says "expand" or "simplify", write every term and check with $x=1$. If it asks for "the coefficient of" or "the term containing" one power, jump straight to that term: decide the power on each part of the binomial first (they must add to $n$), pick the matching entry of row $n$, then multiply the entry, the powers of the numbers and any signs. If the row you need is further down than you remember, build it from the row above using the addition rule; rows 0 to 7 are quick to write out on any test. The same idea also lets you work backwards: if ${(1+x)}^n$ begins $1+8x+\cdots$, the second entry of the row is 8, so $n=8$.