Exercise 7.2 — Q1: Binomial Theorem Expansions

Binomial Theorem

For any positive integer $n$, the Binomial Theorem states:

$(a+b{)}^n={\sum }_{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})a^{n-r}{b}^r$

where the binomial coefficient is:

$\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$

General Term

The general term (the $(r+1)$-th term) of the expansion is:

$T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r,r=0,1,2,\dots ,n$

Total Number of Terms

The expansion of $(a+b{)}^n$ has $n+1$ terms.

Middle Term(s)

• If $n$ is even: one middle term, $T_{\frac{n}{2}+1}$
• If $n$ is odd: two middle terms, $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$

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Pascal's Triangle and Binomial Coefficients

Binomial coefficients can be read directly from Pascal's Triangle:

$\begin{array}{cccccccccc}n=0:& & & & 1& & & & & \\ n=1:& & & 1& & 1& & & & \\ n=2:& & 1& & 2& & 1& & & \\ n=3:& 1& & 3& & 3& & 1& & \\ n=4:& 1& & 4& & 6& & 4& & 1\end{array}$

Each entry is the sum of the two entries directly above it. The $r$-th entry (0-indexed) in row $n$ gives $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$.

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Worked Examples

Example 1: Expand $(x+y{)}^4$

Using the Binomial Theorem with $a=x$, $b=y$, $n=4$:

$(x+y{)}^4=(\genfrac{}{}{0ex}{}{4}{0})x^4+(\genfrac{}{}{0ex}{}{4}{1})x^{3}y+(\genfrac{}{}{0ex}{}{4}{2})x^2{y}^2+(\genfrac{}{}{0ex}{}{4}{3})x{y}^3+(\genfrac{}{}{0ex}{}{4}{4})y^4$

$=x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4$

Example 2: Expand $(2x-3{)}^3$

Here $a=2x$, $b=-3$, $n=3$:

$(2x-3{)}^3=(\genfrac{}{}{0ex}{}{3}{0})(2x{)}^3+\left(\genfrac{}{}{0ex}{}{3}{1}\right)(2x{)}^2(-3)+(\genfrac{}{}{0ex}{}{3}{2})(2x)(-3{)}^2+\left(\genfrac{}{}{0ex}{}{3}{3}\right)(-3{)}^3$

$=8x^3+3(4x^2)(-3)+3(2x)(9)+(-27)$

$=8x^3-36x^2+54x-27$

Example 3: Find the middle term of $(x+2{)}^6$

Here $n=6$ (even), so the middle term is $T_{\frac{6}{2}+1}=T_4$.

$T_4=T_{3+1}=\left(\genfrac{}{}{0ex}{}{6}{3}\right)x^{6-3}(2{)}^3=20\cdot x^3\cdot 8=160x^3$