Pascal's Triangle

The Triangle of Coefficients

Section 3 proved the key identity $^{n}C_{r-1}+{}^{n}C_r={}^{n+1}C_r$: each binomial coefficient of index $n+1$ is the sum of two neighbouring coefficients of index $n$. Arranging the coefficients $^{n}C_0,{}^{n}C_1,\ldots,{}^{n}C_n$ of successive indices $n=0,1,2,\ldots$ in rows, one below the other and centred, produces a triangular array known as Pascal's Triangle:

Index 0:                 1
Index 1:               1   1
Index 2:             1   2   1
Index 3:           1   3   3   1
Index 4:         1   4   6   4   1
Index 5:       1   5  10  10   5   1
Index 6:     1   6  15  20  15   6   1

Construction rule. Every row begins and ends with $1$ (since $^{n}C_0={}^{n}C_n=1$ for every $n$). Every interior entry is the sum of the two entries diagonally above it in the previous row -- this is exactly the identity $^{n}C_{r-1}+{}^{n}C_r={}^{n+1}C_r$ read pictorially. For instance, the entry $15$ in row 6 comes from adding the $5$ and $10$ that flank it in row 5; the entry $20$ in the middle of row 6 comes from adding the two $10$'s that flank it in row 5.

Reading Off an Expansion

Because row $n$ of the triangle is precisely the list of coefficients $^{n}C_0,{}^{n}C_1,\ldots,{}^{n}C_n$ that the binomial theorem needs, an expansion of $(a+b)^n$ can be written down directly once row $n$ is known -- without evaluating a single factorial. For example, row 5 reads $1,5,10,10,5,1$, so

matching what the formula $^{5}C_r$ would give for $r=0,1,2,3,4,5$. …