Pascal's triangle

Algebra: Binomial Coefficient

Pascal's triangle

Binomial coefficients

We have already seen the next sum formulas for powers. \[\begin{array}{rcc} (a+b)^2 & = & a^2+2ab+b^2\\ (a+b)^3 & = & a^3+3a^2b+3ab^2+b^3\\ (a+b)^4 & = & a^4+4a^3b+6a^2b^2+4ab^3+b^4\end{array}\]We have arranged the terms in a systematic order with decreasing powers of \(a\): every next power of \(a\) has an exponent, which is decreased by #1#. The integers in front of the terms of the expansion of \((a+b)^n\) with \(n=0,1,2,3,4,\ldots\) can be arranged similarly. They are called the binomial coefficients.

For \(n=2\) the binomial coefficients are #1#, #2#, and #1#.

For \(n=3\) they are #1#, #3#, #3#, and #1#.

For \(n=4\) they are #1#, #4#, #6#, #4#, and #1#.

Two remarks:

The sequences start and end with #1#. This can be explained because the sequence starts with the coefficient of \(a^n\) which is #1# and ends with the coefficient of \(b^n\), which is #1# as well.
When you reverse the sequence, you end up with the same sequence. This is because #a# and #b# have a systematic role: #(b+a)^n=(a+b)^n#.

Pascal's triangle

One could place the binomial coefficients for various values of \(n\) in the so called Pascal's triangle: \[\begin{array}{ccc}
n = 0: & & \phantom{0}1 \\
n = 1: & & \phantom{0}1 \quad \phantom{0}1 \\
n = 2: & & \phantom{0}1 \quad \phantom{0}2 \quad \phantom{0}1 \\
n = 3: & & \phantom{0}1 \quad \phantom{0}3 \quad \phantom{0}3 \quad \phantom{0}1 \\
n = 4: & & \phantom{0}1 \quad \phantom{0}4 \quad \phantom{0}6 \quad \phantom{0}4\quad \phantom{0}1 \\
n = 5: & & \phantom{0}1 \quad \phantom{0}5 \quad 10 \quad 10 \quad \phantom{0}5 \quad \phantom{0}1 \\
n = 6: & & \phantom{0}1 \quad \phantom{0}6 \quad 15 \quad 20 \quad 15 \quad \phantom{0}6 \quad \phantom{0}1 \\
\cdots & & \cdots
\end{array}\]

The sequence with #n=6# means that \[(a+b)^6=a^6+6a^5b+ 15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6\tiny.\]

Pascal's triangle is constructed by the following rule:

Rule to construct the triangle of Pascal: We place ones along the left and the right edge and the rest of the numbers are sums of the corresponding left- and right- upperneigbor.

We discuss the derivation for #n=6#.\[\begin{array}{rcl}(a+b)^5&=&(a+b)^6\cdot(a+b)\\ &=& \left(a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\right)\cdot(a+b)\\ &=& a^6+5a^5b+10a^4b^2+10a^3b^3+\phantom{1}5a^2b^4+\phantom{5}ab^5\\ && \phantom{a^6}+\phantom{5}a^5b+\phantom{1}5a^4b^2+10a^3b^3+10a^2b^4+5ab^5+b^6\\&&\phantom{myspace} \color{blue}{\text{similar terms taken together}}\\ &=&a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6\end{array}\]The coefficient of #a^2b^4# is created by the addition of the coefficient of #ab^4# from #(a+b)^5# (which turn into #a^2b^4# after multiplication of #ab^4# with #a#) and the coefficient of #a^2b^3# from #(a+b)^5# (which turns into #a^2b^4# after multiplication of #a^2b^3# with #b#): #15 = 5+10#.

Generally the coefficient of #a^kb^{n-k}# in #(a+b)^{n}# is created by the addition of the coefficients of #a^{k-1}b^{n-k}# and #a^{k}b^{n-k-1}# in #(a+b)^{n-1}#. The reason is that the term with #a^kb^{n-k}# of #(a+b)^{n}# is established by the following to terms of #(a+b)^{n-1}#:

the term with #a^{k-1}b^{n-k}#, by multiplication with #a# (originating from #(a+b)#) and
the term with #a^{k}b^{n-k-1}#, by multiplication with #b# (originating from the factor #(a+b)#).