Binomial Theorem
Binomial Expansions
Binomial expansions, that is, binomials or other two-term quantities raised to integer powers, are of frequent occurrence. If the general binomial expression is $ a + b $ , then the first few powers are given by:
\[ (a+b)^{0}=1 \]
\[ (a+b)^{1}=a+b \]
\[ (a+b)^{2}=a^{2}+2ab+b^{2} \]
\[ (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} \]
Patterns in Binomial Expansions
Many patterns have been observed in the sequence of expansions of $ (a + b)^{n} $ . For example:
There are $ n + 1 $ terms in the expansion of $ (a + b)^{n} $ .
The exponent of a starts in the first term as n, and decreases by 1 in each succeeding term down to 0 in the last term.
The exponent of b starts in the first term as 0, and increases by 1 in each succeeding term up to n in the last term.
Binomial Theorem
The binomial theorem gives the expansion of $ (a + b)^{n} $ . In its most compact form, this is written as follows:
\[ (a+b)^{n}=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r} \]
The symbols $ $ are called the binomial coefficients, defined as: $ = $ .
EXAMPLE 44.1 Calculate the binomial coefficients $ $ and verify the expansion of $ (a + b)^{3} $ above.
\[ \binom{3}{0}=\frac{3!}{0!(3-0)!}=\frac{3!}{1\cdot3!}=1 \]
\[ \binom{3}{1}=\frac{3!}{1!(3-1)!}=\frac{3!}{1!2!}=\frac{3\cdot2\cdot1}{1(2\cdot1)}=3 \]
\[ \binom{3}{2}=\frac{3!}{2!(3-2)!}=\frac{3!}{2!1!}=\frac{3\cdot2\cdot1}{(2\cdot1)1}=3 \]
\[ \binom{3}{3}=\frac{3!}{3!(3-3)!}=\frac{3!}{3!0!}=\frac{3!}{3!\cdot1}=1 \]
Therefore
\[ \begin{aligned}(a+b)^{3}&=\sum_{r=0}^{3}\binom{3}{r}a^{3-r}b^{r}=\binom{3}{0}a^{3-0}b^{0}+\binom{3}{1}a^{3-1}b^{1}+\binom{3}{2}a^{3-2}b^{2}+\binom{3}{3}a^{3-3}b^{3}\\&=1a^{3}b^{0}+3a^{2}b^{1}+3a^{1}b^{2}+1a^{0}b^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\end{aligned} \]
Properties of the Binomial Coefficients
The following are readily verifiable:
\[ \binom{n}{0}=\binom{n}{n}=1 \]
\[ \binom{n}{r}=\binom{n}{n-r} \]
\[ \binom{k}{r-1}+\binom{k}{r}=\binom{k+1}{r} \]
The binomial coefficients are also referred to as the combinatorial symbols. Then the designation $ {n}C{r} $ is used, with
\[ {}_{n}C_{r}={n\choose r} \]
Finding Particular Terms of a Binomial Expansion
In the binomial expansion of $ (a + b)^{n} $ , r, the index of the terms, starts at 0 in the first term and goes up to n in the $ n + 1 $ st term. Thus the index r is equal to j - 1 in the jth term. If a particular term is desired, it is generally thought of as the $ j + 1 $ st term; then r is equal to j and the value of the $ j + 1 $ st term is given by
\[ \binom{n}{j}a^{n-j}b^{j} \]
EXAMPLE 44.2 Find the fifth term in the expansion of $ (a + b)^{16} $
Here n = 16 and $ j + 1 = 5 $ , thus j = 4 and the term is given by
\[ {n\choose j}a^{n-j}b^{j}={16\choose4}a^{16-4}b^{4}=\frac{16!}{4!(16-4)!}a^{16-4}b^{4}=1820a^{12}b^{4} \]
SOLVED PROBLEMS
44.1. Calculate the binomial coefficients:
\[ \left(a\right)\binom{4}{2};\left(b\right)\binom{8}{5};\left(c\right)\binom{12}{1};\left(d\right)\binom{n}{n-1} \]
\[ \left(a\right)\binom{4}{2}=\frac{4!}{2!(4-2)!}=\frac{4!}{2!2!}=\frac{4\cdot3\cdot2\cdot1}{2\cdot1\cdot2\cdot1}=6 \]
\[ \left(b\right)\binom{8}{5}=\frac{8!}{5!(8-5)!}=\frac{8!}{5!3!}=\frac{8\cdot7\cdot6\cdot5!}{5!(3\cdot2\cdot1)}=\frac{8\cdot7\cdot6}{3\cdot2\cdot1}=56 \]
\[ \left(\mathrm{c}\right)\binom{12}{1}=\frac{12!}{1!(12-1)!}=\frac{12!}{1!11!}=\frac{12\cdot11!}{1\cdot11!}=12 \]
\[ \left(\mathrm{d}\right)\binom{n}{n-1}=\frac{n!}{(n-1)![n-(n-1)]!}=\frac{n(n-1)!}{(n-1)!1!}=n \]
44.2. Show that $ = = 1 $ .
\[ \binom{n}{n}=\frac{n!}{n!(n-n)!}=\frac{n!}{n!0!}=\frac{n!}{n!(1)}=1.Similarly\binom{n}{0}=\frac{n!}{0!(n-0)!}=\frac{n!}{1(n!)}=1 \]
44.3. Show that $ = = $ for any integer r < n.
Note that $ n! = n(n - 1)(n - 2) (r + 1)r $ . Hence
\[ \begin{aligned}\binom{n}{r}&=\frac{n!}{r!(n-r)!}=\frac{n(n-1)(n-2)\cdot\cdots\cdot(r+1)r\cdot\cdots\cdot1}{r!(n-r)!}=\frac{n(n-1)(n-2)\cdot\cdots\cdot(r+1)r!}{r!(n-r)!}\\&=\frac{n(n-1)\cdot\cdots\cdot(r+1)}{(n-r)!}\end{aligned} \]
Similarly, $ n! = n(n - 1)(n - 2) (n - r + 1)(n - r) $ . Hence
\[ \begin{aligned}\binom{n}{r}&=\frac{n!}{r!(n-r)!}=\frac{n(n-1)(n-2)\cdot\cdots\cdot(n-r+1)(n-r)\cdot\cdots\cdot1}{r!(n-r)!}\\&=\frac{n(n-1)(n-2)\cdot\cdots\cdot(n-r+1)(n-r)!}{r!(n-r)!}\\&=\frac{n(n-1)\cdot\cdots\cdot(n-r+1)}{r!}\end{aligned} \]
44.4. Use the results of the previous problems to write out the terms of $ (a + b)^{4} $ .
\[ \begin{aligned}(a+b)^{4}&=\sum_{r=0}^{4}\binom{4}{r}a^{4-r}b^{r}\\&=\binom{4}{0}a^{4-0}b^{0}+\binom{4}{1}a^{4-1}b^{1}+\binom{4}{2}a^{4-2}b^{2}+\binom{4}{3}a^{4-3}b^{3}+\binom{4}{4}a^{4-4}b^{4}\\&=1a^{4}+\frac{4}{1}a^{3}b+\frac{4\cdot3}{2\cdot1}a^{2}b^{2}+\frac{4\cdot3\cdot2}{3\cdot2\cdot1}a^{1}b^{3}+1b^{4}\\&=a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4}\end{aligned} \]
44.5. Write the binomial expansion of $ (3x - 5y)^{4} $
Use the result of the previous problem with a = 3x and b = -5y. Then
\[ \begin{aligned}[(3x)+(-5y)]^{4}&=(3x)^{4}+4(3x)^{3}(-5y)+6(3x)^{2}(-5y)^{2}+4(3x)(-5y)^{3}+(-5y)^{4}\\&=81x^{4}-540x^{3}y+1350x^{2}y^{2}-1500xy^{3}+625y^{4}\end{aligned} \]
44.6. Write the first three terms in the binomial expansion of $ (a + b)^{20} $
Since $ (a + b)^{20} = _{r=0}^{20} a^{20 - r} b^{r} $ , the first three terms can be written as
\[ \begin{align*}\binom{20}{0}a^{20-0}b^{0}+\binom{20}{1}a^{20-1}b^{1}+\binom{20}{2}a^{20-2}b^{2}&=1a^{20}+\frac{20}{1}a^{19}b+\frac{20\cdot19}{2\cdot1}a^{18}b^{2}\\&=a^{20}+20a^{19}b+190a^{18}b^{2}\end{align*} \]
44.7. Write the first three terms in the binomial expansion of $ (2x^{5} + 3t{2}){12} $ .
Take \(a=2x^{5}\) and \(b=3t^{2}\). Then \((2x^{5}+3t^{2})^{12}=\sum_{r=0}^{12}\binom{12}{r}(2x^{5})^{12-r}(3t^{2})^{r}\). The first three terms of this can be written as
\[ \begin{aligned}\binom{12}{0}&(2x^{5})^{12}+\binom{12}{1}(2x^{5})^{11}(3t^{2})+\binom{12}{2}(2x^{5})^{10}(3t^{2})^{2}\\&=4096x^{60}+12(2048x^{55})(3t^{2})+\frac{12\cdot11}{2\cdot1}(1024x^{50})(9t^{4})=4096x^{60}+73,728x^{55}t^{2}+608,256x^{50}t^{4}\end{aligned} \]
44.8. Show that $ = $ .
Substitute \(n - r\) for \(r\) in the definition of \(\binom{n}{r}\). Then
\[ \binom{n}{n-r}=\frac{n!}{(n-r)![n-(n-r)]!}=\frac{n!}{(n-r)!r!}=\frac{n!}{r!(n-r)!}=\binom{n}{r} \]
44.9. Show that $ + = $ .
Note first that \(r! = r(r - 1)!\). Also, \((k + 1)! = (k + 1)k!\) and \((k - r + 1)! = (k - r + 1)(k - r)!\). Then
\[ \binom{k}{r-1}+\binom{k}{r}=\frac{k!}{(r-1)!(k-r+1)!}+\frac{k!}{r!(k-r)!} \]
The LCD for the two fractional expressions on the right is $ r!(k - r + 1)! $ . Rewriting with this common denominator yields:
\[ \begin{aligned}\frac{k!}{(r-1)!(k-r+1)!}+\frac{k!}{r!(k-r)!}&=\frac{rk!}{r(r-1)!(k-r+1)!}+\frac{(k-r+1)k!}{(k-r+1)r!(k-r)!}\\&=\frac{rk!}{r!(k-r+1)!}+\frac{(k-r+1)k!}{r!(k-r+1)!}\end{aligned} \]
The two expressions on the right can be combined to yield:
\[ \frac{rk!}{r!(k-r+1)!}+\frac{(k-r+1)k!}{r!(k-r+1)!}=\frac{rk!+(k-r+1)k!}{r!(k-r+1)!}=\frac{(r+k-r+1)k!}{r!(k-r+1)!}=\frac{(k+1)k!}{r!(k-r+1)!} \]
The last expression is precisely
\[ \frac{(k+1)!}{r!(k+1-r)!}=\binom{k+1}{r} \]
44.10. Show that the binomial coefficients can be arranged in the form shown in Fig. 44-1.
\[ \begin{array}{r} 1 \\ 1 \phantom{0} 1 \\ 1 2 \phantom{1} 1 \\ 1 3 \phantom{3} 1 \\ 1 4 \phantom{6} 4 \phantom{1} \\ \hline \end{array} \]
where each entry, except the 1’s, is the sum of the two entries above it and to the right and left. (This triangular arrangement is often called Pascal’s triangle.)
Clearly the first two rows represent $ (a + b)^{0} = 1 $ and the coefficients of $ (a + b)^{1} = 1a + 1b $ . For the other rows, note that the first and the last binomial coefficients in each are given by
\[ \binom{n}{0}=1\qquad and\qquad\binom{n}{n}=1 \]
respectively. For all other coefficients, since, as proved in the previous problem,
\[ \binom{k}{r-1}+\binom{k}{r}=\binom{k+1}{r} \]
each entry in the $ k + 1 $ st row is the sum of the two entries in the kth row above it and to the right and left.
44.11. Use mathematical induction to prove the binomial theorem for positive integers n
Let $ P_{n} $ be the statement of the binomial theorem:
\[ (a+b)^{n}=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r} \]
Then $ P_{1} $ is the statement $ (a + b)^{1} = _{r=0}^{1} a^{1-r} b^{r} $ .
$ P_{k} $ is the statement $ (a + b)^{k} = _{r=0}^{k} a^{k-r} b^{r} $
$ P_{k+1} $ is the statement $ (a + b)^{k+1} = _{r=0}^{k+1} a^{k+1-r} b^{r} $
Now $ P_{1} $ is true, since the left side is $ a + b $ and the right side is
\[ {1\choose0}a^{1-0}b^{0}+{1\choose1}a^{1-1}b^{1}=1a+1b=a+b \]
Assume the truth of $ P_{k} $ , and, comparing it with $ P_{k+1} $ , note that the left side of $ P_{k+1} $ is $ a + b $ times the left side of $ P_{k} $ . Hence, starting with $ P_{k} $ , multiply both sides by $ a + b $ :
\[ (a+b)(a+b)^{k}=(a+b)\sum_{r=0}^{k}\binom{k}{r}a^{k-r}b^{r} \]
\[ \begin{align*}(a+b)^{k+1}&=a\sum_{r=0}^{k}\binom{k}{r}a^{k-r}b^{r}+b\sum_{r=0}^{k}\binom{k}{r}a^{k-r}b^{r}\\&=\sum_{r=0}^{k}\binom{k}{r}a^{k+1-r}b^{r}+\sum_{r=0}^{k}\binom{k}{r}a^{k-r}b^{r+1}\end{align*} \]
Writing out the terms of the sums yields:
\[ \begin{aligned}&\binom{k}{0}a^{k+1}+\binom{k}{1}a^{k}b+\binom{k}{2}a^{k-1}b^{2}+\cdots+\binom{k}{k-1}a^{2}b^{k-1}+\binom{k}{k}a b^{k}\\ &+\binom{k}{0}a^{k}b+\binom{k}{1}a^{k-1}b^{2}+\cdots+\binom{k}{k-2}a^{2}b^{k-1}+\binom{k}{k-1}a b^{k}+\binom{k}{k}b^{k+1}\\ \end{aligned} \]
Combining like terms, and noting that $ = 1 = $ and $ = 1 = $ , yields
\[ \begin{aligned}\binom{k+1}{0}a^{k+1}&+\binom{\binom{k}{1}+\binom{k}{0}}{}\binom{a^{k}b}{}+\binom{\binom{k}{2}+\binom{k}{1}}{}\binom{a^{k-1}b^{2}+\cdots}{}\\&+\binom{\binom{k}{k-1}+\binom{k}{k-2}}{}\binom{a^{2}b^{k-1}+\binom{\binom{k}{k}+\binom{k}{k-1}}{}}{}ab^{k}+\binom{k+1}{k+1}b^{k+1}\\ =&\binom{k+1}{0}a^{k+1}+\binom{k+1}{1}a^{k}b+\binom{k+1}{2}a^{k-1}b^{2}+\cdots+\binom{k+1}{k-1}a^{2b^{k-1}}+\binom{k+1}{k}ab^{k}+\binom{k+1}{k+1}b^{k+1}\end{aligned} \]
Thus, writing the last expression in summation notation,
\[ (a+b)^{k+1}=\sum_{r=0}^{k+1}\binom{k+1}{r}a^{k+1-r}b^{r} \]
But this is precisely the statement $ P_{k+1} $ . Thus the truth of $ P_{k+1} $ follows from the truth of $ P_{k} $ . Thus, by the principle of mathematical induction, $ P_{n} $ , that is, the binomial theorem, holds for all positive integers n.
44.12. Write the eighth term in the expansion of $ (+)^{13} $ .
The \((j+1)\)st term in the expansion of \((a+b)^{n}\) is given by \(\binom{n}{j}a^{n-j}b^{j}\). Here \(n=13\) and \(j+1=8\); hence \(j=7\). Thus the required term is
\[ {13\choose7}(\sqrt{x})^{13-7}\bigg(\frac{1}{\sqrt{x}}\bigg)^{7}=\frac{13!}{(13-7)!7!}\frac{(\sqrt{x})^{6}}{(\sqrt{x})^{7}}=\frac{13\cdot12\cdot11\cdot10\cdot9\cdot8}{6!\sqrt{x}}=\frac{1716}{\sqrt{x}} \]
44.13. Use the binomial theorem to approximate $ (1.01)^{20} $ to three decimal places.
Expand $ (1 + 0.01)^{20} $ to obtain:
\[ \begin{aligned}\binom{20}{0}&1^{20}+\binom{20}{1}1^{19}(0.01)^{1}+\binom{20}{2}1^{18}(0.01)^{2}+\binom{20}{3}1^{17}(0.01)^{3}+\binom{20}{4}1^{16}(0.01)^{4}+\cdots\\&=1+20(0.01)+\frac{20\cdot19}{2\cdot1}(0.0001)+\frac{20\cdot19\cdot18}{3\cdot2\cdot1}(0.000001)+\frac{20\cdot19\cdot18\cdot17}{4\cdot3\cdot2\cdot1}(10^{-8})+\cdots\\&=1+0.2+0.019+0.00114+0.00004845+\cdots\\&=1.220188\cdots\end{aligned} \]
where the neglected terms have no effect on the third decimal place. Thus $ (1.01)^{20} $ to three decimal places.
SUPPLEMENTARY PROBLEMS
44.14. Calculate the binomial coefficients: (a) $ $ ; (b) $ $ ; (c) $ $ ; (d) $ $
Ans. (a) 15; (b) 28; (c) 220; (d) $ $
44.15. Write the binomial expansion of (a) $ (a + b)^{5} $ ; (b) $ (2x + y)^{5} $ .
\[ Ans.\quad(a)a^{5}+5a^{4}b+10a^{3}b^{2}+10a^{2}b^{3}+5ab^{4}+b^{5};(b)32x^{5}+80x^{4}y+80x^{3}y^{2}+40x^{2}y^{3}+10xy^{4}+y^{5} \]
44.16. Write the binomial expansion of (a) $ (4s - 3t)^{3} $ ; (b) $ (2a - )^{5} $ .
Ans. (a) $ 64s{3}-144s{2}t+108st{2}-27t{3} $ ; (b) $ 32a{5}-16a{4}b+a{3}b{2}-a{2}b{3}+ab{4}-\frac{b{5}}{3125} $
44.17. Prove that $ + + + + = 2^{n} $ , that is, that the sum of the binomial coefficients for any power n is equal to $ 2^{n} $ . [Hint: Consider the binomial expansion of $ (1 + 1)^{n} $ .]
44.18. Find the middle term in the binomial expansion of (a) $ (3x - )^{14} $ ; (b) $ (x^{3} + 2y{3}){10} $ .
Ans. (a) $ -3432x{7}y{7} $ ; (b) $ 8064x{15}y{15} $
44.19. It is shown in calculus that if $ |x| < 1 $ and $ $ is not a positive integer, then $ (1 + x)^{} = _{j=0}^{} x^{j} $ with $ = $ . Use this formula to write the first three terms of the binomial expansion of (a) $ (1 + x)^{-2} $ ; (b) $ (1 + x)^{1/2} $ .
Ans. (a) $ 1 - 2x + 3x^{2} $ ; (b) $ 1 + x - x^{2} $