In the previous section,
we completed a discussion on General and Middle terms in binomial expansions. In this
section, we will see some miscellaneous examples.
Solved example 8.10
Find the term independent of x in the expansion of $\left({\frac{3}{2}}x^2~-~\frac{1}{3x} \right)^6$.
Solution:
1. Assume that the term independent of x occurs in the (r+1)th term.
• We know that, the (r+1)th term of the binomial expansion (a-b)n is given by:
nCr (-1)r an-r br
• In our present case, n = 6, $a={\frac{3}{2}}x^2,~~b=\frac{1}{3x}$
2. So we can write:
(r+1)th term of the expansion of $\left({\frac{3}{2}}x^2~-~\frac{1}{3x} \right)^6~=~{}^{6} {\rm{C}}_r\;(-1)^r \; \left({\frac{3}{2}}x^2 \right)^{6-r} \;\left(\frac{1}{3x} \right)^r$
This can be simplified as:
${}^{6}
{\rm{C}}_r\;(-1)^r \; \left({\frac{3}{2}}\right)^{6-r}\; \left(x^2 \right)^{6-r}
\;\left(\frac{1}{3^r x^r} \right)~=~{}^{6}
{\rm{C}}_r\;(-1)^r \; \left({\frac{3}{2}}\right)^{6-r}\; \left(x^{12-2r} \right)\;\left(\frac{1}{3^r x^r} \right)$
3. Let us compare the indices of x:
♦ The index of x in the numerator is 12-2r
♦ The index of x in the denominator is r
• These two indices must be equal. So we get: 12 - 2r = r
⇒ 12 = 3r, which gives: r = 4
4. So we can write:
The term with r = 4, will be independent of x. That means, the fifth term will be independent of x.
• From (2), we get:
(4+1)th term = ${}^{6}
{\rm{C}}_4\;(-1)^4 \; \left({\frac{3}{2}}x^2 \right)^{6-4}
\;\left(\frac{1}{3x} \right)^4~=~{}^{6}
{\rm{C}}_4\; × \;1 \; × \; \left({\frac{3}{2}}x^2 \right)^{2}
\;\left(\frac{1}{3^4 x^4} \right)$
$~=~15 × \; \left({\frac{3^2}{2^2}}\right)\; × x^4
\; × \;\left(\frac{1}{3^4 x^4} \right)~=~\frac{15}{36}~=~\frac{5}{12}$
Solved example 8.11
If the coefficients of ar-1, ar and ar+1 in the expansion of (1 + a)n are in arithmetic progression, prove that n2 – n(4r + 1) + 4r2 – 2 = 0.
Solution:
1. We know that, the (r+1)th term of the binomial expansion (a+b)n is given by:
nCr an-r br
• In our present case, n = n, a = 1 and b = a
2. So we can write:
(r+1)th
term of the expansion of (1 + a)n is: nCr 1n-r ar
= nCr ar
3. It follows that:
• rth
term of the expansion of (1 + a)n will be: nCr-1 ar-1
• (r+2)th
term of the expansion of (1 + a)n will be: nCr+1 ar+1
4. So the three coefficients in arithmetic progression are:
nCr-1, nCr, nCr+1
• Since they are in arithmetic progression, the common difference will be the same. We can write:
nCr - nCr-1 = nCr+1 - nCr
⇒ nCr-1 + nCr+1 = 2 × nCr
• This can be simplified as:
$\begin{array}{ll}
{}&{}
&{\frac{n!}{(r-1)!(n-r+1)!}~+~\frac{n!}{(r+1)!(n-r-1)!}}& {}={}
&{2 × \frac{n!}{r!(n-r)!}}& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{\frac{n!}{(r-1)!(n-r+1)(n-r)(n-r-1)!}~+~\frac{n!}{(r+1)r(r-1)!(n-r-1)!}}& {}={}
&2 × \frac{n!}{r!(n-r)!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{\frac{n!r(r+1)~+~n!(n-r+1)(n-r)}{(r-1)!(n-r+1)(n-r)(n-r-1)!r (r+1)}}& {}={}
&2 × \frac{n!}{r!(n-r)!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{\frac{n!r(r+1)~+~n!(n-r+1)(n-r)}{[r(r+1)(r-1)!][(n-r+1)(n-r)(n-r-1)!]}}& {}={}
&2 × \frac{n!}{r!(n-r)!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{\frac{n!r(r+1)~+~n!(n-r+1)(n-r)}{[(r+1)!][(n-r+1)(n-r)!]}}& {}={}
&2 × \frac{n!}{r!(n-r)!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}
&\color {green} {\text{[n! and (n-r)! can be cancelled from both sides]}}& {}&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{\frac{r(r+1)~+~(n-r+1)(n-r)}{[(r+1)!][(n-r+1)]}}& {}={}
&2 × \frac{1}{r!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{\frac{r(r+1)~+~(n-r+1)(n-r)}{[(r+1)r!][(n-r+1)]}}& {}={}
&2 × \frac{1}{r!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}
&\color {green} {\text{[r! can be cancelled from both sides]}}& {}&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{\frac{r(r+1)~+~(n-r+1)(n-r)}{[(r+1)][(n-r+1)]}}& {}={}
&2 × 1& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{r^2+r+n^2-nr-nr+r^2+n-r}& {}={}
&2 × [nr-r^2+r+n-r+1]& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{2r^2+n^2-2nr+n}& {}={}
&2nr-2r^2+2n+2& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{4r^2+n^2-4nr-n-2}& {}={}
&0& {}
&{}& {}
&{}& {}&{}& {} &{} \\
{}&{}\Rightarrow{}
&{n^2-n(4r+1)+4r^2-2}& {}={}
&0& {}
&{}& {}
&{}& {}&{}& {} &{} \\
\end{array}$
Solved example 8.12
Show that the coefficient of the middle term in the expansion of (1 + x)2n is equal to the sum of the coefficients of two middle terms in the expansion of (1 + x)2n-1.
Solution:
1. Consider (1 + x)2n. The index will be always even. So there will always be an odd number of terms. As a result, there will be a single middle term.
• The position of that single middle term is: $\frac{n}{2}~+~1$
• In our present case, n = 2n.
• So position of the middle term is $\frac{2n}{2}~+~1~=~(n+1)$
• We know that, the (r+1)th term of the binomial expansion (a+b)n is given by:
nCr an-r br
• In our present case, n = 2n, a = 1 and b = x
• So we can write:
(r+1)th
term of the expansion of (1 + x)2n is: 2nCr 12n-r ar
= 2nCr ar
• Based on this, we can write:
Middle term, which is the (n+1)th term will be: 2nCn an
• Coefficient of this middle term is 2nCn
• This can be simplified as:
$\frac{(2n)!}{n!(2n-n)!}~=~\frac{2n(2n-1)!}{n!n!}~=~\frac{2n(2n-1)!}{n!n(n-1)!}~=~\frac{2(2n-1)!}{n!(n-1)!}$
2. Consider (1 + x)2n-1. The index will be always odd. So
there will always be an even number of terms. As a result, there will be a
two middle terms.
• The positions of those middle terms are: $\frac{n+1}{2}~\text{and}~\frac{n+1}{2}~+~1$
• In our present case, n = 2n-1.
• So position of the middle terms are $\frac{(2n-1)+1}{2}~\text{and}~\frac{(2n-1)+1}{2}~+~1$
• So the positions are: n and (n+1)
• We know that, the (r+1)th term of the binomial expansion (a+b)n is given by:
nCr an-r br
• In our present case, n = 2n-1, a = 1 and b = x
• So we can write:
(r+1)th
term of the expansion of (1 + x)2n-1 is: 2n-1Cr 12n-1-r ar
= 2n-1Cr ar
• Based on this, we can write:
♦ The second middle term, which is the (n+1)th term will be: 2n-1Cn an
♦ The first middle term, which is the nth term will be: 2n-1Cn-1 an-1
• Sum of the coefficients of these two middle terms = 2n-1Cn + 2n-1Cn-1
This can be simplified as:
$\frac{(2n-1)!}{n!(2n-1-n)!}~+~\frac{(2n-1)!}{(n-1)!(2n-1-n+1)!}$
= $\frac{(2n-1)!}{n!(n-1)!}~+~\frac{(2n-1)!}{(n-1)!n!}$
= $\frac{2(2n-1)!}{n!(n-1)!}$
3. Comparing the results in (1), and (2), we see that:
Result in (2) is same as the result in (1)
Solved example 8.13
Find the coefficient of a4 in the product (1 + 2a)4 (2 – a)5 using binomial
theorem.
Solution:
1. Let us expand each of the factors of the given product. We get:
• First factor $(1+2a)^4$
$\begin{array}{ll}
{}={}&{}^4 {\rm{C}}_0 × 1^{4-0} × (2a)^0
&{}+{}& {}^4
{\rm{C}}_1 × 1^{4-1} × (2a)^1
&{}+{}& {}^4 {\rm{C}}_2 × 1^{4-2} × (2a)^2
&{}+{}& {}^4
{\rm{C}}_3 × 1^{4-3} × (2a)^3
&{}+{}& {}^4
{\rm{C}}_4 × 1^{4-4} × (2a)^4 \\
{}={}&{}^4 {\rm{C}}_0 × 1 × 1
&{}+{}& {}^4 {\rm{C}}_1 ×
1 × (2a)
&{}+{}& {}^4 {\rm{C}}_2 × 1 × 4a^2
&{}+{}& {}^4 {\rm{C}}_3 × 1 × 8a^3
&{}+{}& {}^4 {\rm{C}}_4 × 1 × 16a^4 \\
{}={}&1
&{}+{}&8a
&{}+{}& 24a^2
&{}+{}& 32a^3
&{}+{}& 16a^4 \\
\end{array}$
• Second factor $(2-a)^5$
$\begin{array}{ll}
{}={}&{}^5 {\rm{C}}_0 × (-1)^0 × 2^{5-0} × a^0
&{}+{}& {}^5 {\rm{C}}_1 × (-1)^1 × 2^{5-1} × a^1
&{}+{}& {}^5 {\rm{C}}_2 × (-1)^2 × 2^{5-2} × a^2
&{}+{}& {}^5 {\rm{C}}_3 × (-1)^3 × 2^{5-3} × a^3
&{}+{}& {}^5 {\rm{C}}_4 × (-1)^4 × 2^{5-4} × a^4
&{}+{}& {}^5 {\rm{C}}_5 × (-1)^5 × 2^{5-5} × a^5 \\
{}={}&{}^5 {\rm{C}}_0 × 2^5 × 1
&{}-{}& {}^5 {\rm{C}}_1 × 2^4 × a
&{}+{}& {}^5 {\rm{C}}_2 × 2^3 × a^2
&{}-{}& {}^5 {\rm{C}}_3 × 2^2 × a^3
&{}+{}& {}^5 {\rm{C}}_4 × 2^1 × a^4
&{}-{}& {}^5 {\rm{C}}_5 × 2^0 × a^5 \\
{}={}&1 × 2^5 × 1
&{}-{}& 5 × 2^4 × a
&{}+{}& 10 × 2^3 × a^2
&{}-{}& 10 × 2^2 × a^3
&{}+{}& 5 × 2^1 × a^4
&{}-{}& 1 × 2^0 × a^5 \\
{}={}&32
&{}-{}& 80a
&{}+{}& 80a^2
&{}-{}& 40a^3
&{}+{}& 10a^4
&{}-{}& a^5 \\
\end{array}$
2. So we can write:
(1 + 2a)4 × (2 – a)5 =
$\left(1+8a+24a^2+32a^3+16a^4 \right) × \left(32-80a+80a^2-40a^3+10a^4-a^5 \right)$
3. There is no need to do the actual multiplication. There is an easier method to find just the required terms. It can be explained in 4 steps:
(i) Pick each term from the first factor.
(ii) Try to pair it with each term of the second factor.
(iii) All the pairs which give a4 are to be selected.
(iv) All the pairs which do not give a4 are to be discarded.
4. Thus we get:
$[1 × 10a^4] + [8a × -40a^3]+[24a^2 × 80a^2]+[32a^3 × -80a]+[16a^4 × 32]$
= $[1 × 10a^4] - [8a × 40a^3]+[24a^2 × 80a^2]-[32a^3 × 80a]+[16a^4 × 32]$
= $10a^4-320a^4+1920a^4-2560a^4+512a^4$
= $-438a^4$
• So the coefficient of a4 in the given product is -438
Solved example 8.14
Find the rth term from the end in the expansion of (x + a)n.
Solution:
1. We know that, the (r+1)th term of the binomial expansion (a+b)n is given by:
nCr an-r br
• In our present case, a = x and b = a.
• So we get:
(r+1)th term of the binomial expansion (x+a)n is: nCr xn-r ar
2. In the expansion , there will be (n+1) terms. That means, the (n+1)th term will be the last term.
• Using the result in (1), we can write:
The last term, which is the (n+1)th term will be nCn xn-n an
3. But the last term, is the first term from the end. So we can write:
First term from the end will be nCn xn-n an
4. Now, the nth term will be the second term from the end.
• Using the result in (1), we can write:
The second term from the end, which is the nth term will be nCn-1 xn-(n-1) an-1
5. similarly, the (n-1)th term will be the third term from the end.
• Using the result in (1), we can write:
The third term from the end, which is the (n-1)th term will be nCn-2 xn-(n-2) an-2
6. similarly, the (n-2)th term will be the fourth term from the end.
• Using the result in (1), we can write:
The fourth term from the end, which is the (n-2)th term will be nCn-3 xn-(n-3) an-3
7. Thus we see a pattern. Based on that pattern, we can write:
The rth term from the end will be nCn-(r-1) xn-[n-(r-1)] an-(r-1)
• Simplifying this, we get: nCn-r+1 xn-n+r-1 an-r+1
= nCn-r+1 xr-1 an-r+1
Solved example 8.15
Find the term independent of x in the expansion of $\left(\sqrt[3]{x}~+~\frac{1}{2 \sqrt[3]{x}} \right)^{18},~x>0$.
Solution:
1. We know that, the (r+1)th term of the binomial expansion (a+b)n is given by:
nCr an-r br
• In our present case, $a~=~\sqrt[3]{x}~~\text{and}~~b~=~\frac{1}{2 \sqrt[3]{x}}$.
• So we get:
(r+1)th term of the given binomial expansion is:
${}^{n} {\rm{C}}_{r}~\left(\sqrt[3]{x} \right)^{n-r}~\left(\frac{1}{2 \sqrt[3]{x}} \right)^r$
2. This can be simplified as:
${}^{n} {\rm{C}}_{r}~\left(x^{\frac{1}{3}} \right)^{n-r}~\left(\frac{1}{2} \right)^r~\left(\frac{1}{x^{\frac{1}{3}}} \right)^r$
= ${}^{n} {\rm{C}}_{r}~\left(x \right)^{^{\frac{n-r}{3}}}~\left(\frac{1}{2} \right)^r~\left(\frac{1}{x^{\frac{r}{3}}} \right)$
3. Now we compare the powers:
♦ Power of x in the numerator
♦ must be equal to
♦ Power of x in the denominator
• Thus we get: $\frac{n-r}{3}~=~\frac{r}{3}$
⇒ n-r = r
⇒ $r~=~\frac{n}{2}~=~\frac{18}{2}~=~9$
4. So we can write:
The term with r = 9, will be independent of x.
That means, the 10th term will be independent of x.
5. From the result in (2), we get:
$T_{9+1}~=~{}^{18} {\rm{C}}_{9}~\left(x \right)^{^{\frac{18-9}{3}}}~\left(\frac{1}{2} \right)^9~\left(\frac{1}{x^{\frac{9}{3}}} \right)~=~{}^{18} {\rm{C}}_{9}~\left(x \right)^{^{\frac{9}{3}}}~\left(\frac{1}{2} \right)^9~\left(\frac{1}{x^{\frac{9}{3}}} \right)~=~{}^{18} {\rm{C}}_{9}~\left(\frac{1}{2} \right)^9~=~{}^{18} {\rm{C}}_{9}~\left(\frac{1}{2^9} \right)$
Solved example 8.16
The sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^2} \right)^m,~x \ne 0$, m being a natural number, is 559. Find the term of the expansion containing x3 .
Solution:
1. We know that, the (r+1)th term of the binomial expansion (a-b)n is given by:
nCr (-1)r an-r br
• In our present case, $n~=~m,~a~=~x~~\text{and}~~b~=~\frac{3}{x^2}$.
• So we get:
(r+1)th term of the given binomial expansion is:
${}^{m} {\rm{C}}_{r}~(-1)^r~\left(x \right)^{m-r}~\left(\frac{3}{x^2} \right)^r$
2. Now we can write the terms:
◼ For the first term, we put r = 0. So we get:
$T_{0+1}~=~{}^{m} {\rm{C}}_{0}~(-1)^0~\left(x \right)^{m-0}~\left(\frac{3}{x^2} \right)^0~=~{}^{m} {\rm{C}}_{0}~ × 1 × x^m × 1~=~\left[{}^{m} {\rm{C}}_{0}\right]\;\left[x^m \right]$
◼ For the second term, we put r = 1. So we get:
$T_{1+1}~=~{}^{m} {\rm{C}}_{1}~ (-1)^1\left(x \right)^{m-1}~\left(\frac{3}{x^2} \right)^1~=~{}^{m} {\rm{C}}_{1}~ × -1 × x^{m-1}~\left(\frac{3^1}{x^{2 × 1}} \right)~=~-\left[{}^{m} {\rm{C}}_{1}\;3^1 \right]\;\left[\frac{x^{m-1}}{x^{2 × 1}}\right]~=~-\left[{}^{m} {\rm{C}}_{1} × 3 \right]\;\left[\frac{x^{m-1}}{x^{2}}\right]$
◼ For the third term, we put r = 2. So we get:
$T_{2+1}~=~{}^{m} {\rm{C}}_{2}~(-1)^2~\left(x \right)^{m-2}~\left(\frac{3}{x^2} \right)^2~=~{}^{m} {\rm{C}}_{2}~ × 1 × ~x^{m-2}~\left(\frac{3^2}{x^{2 × 2}} \right)~=~\left[{}^{m} {\rm{C}}_{2}\;3^2 \right]\;\left[\frac{x^{m-2}}{x^{2 × 2}}\right]~=~\left[{}^{m} {\rm{C}}_{2} × 3^2 \right]\;\left[\frac{x^{m-2}}{x^{4}}\right]$
3. So the coefficients are: $\left[{}^{m} {\rm{C}}_{0}\right],~-\left[{}^{m} {\rm{C}}_{1} × 3 \right],~~\text{and}~~\left[{}^{m} {\rm{C}}_{2} × 3^2 \right]$
We can write:
Sum of the coefficients = $\left[{}^{m} {\rm{C}}_{0}\right]~-~\left[{}^{m} {\rm{C}}_{1} × 3 \right]~+~\left[{}^{m} {\rm{C}}_{2} × 3^2 \right]~=~559$
4. This can be simplified as:
$\frac{m!}{0!(m-0)!}~-~\frac{3 × m!}{1!(m-1)!}~+~\frac{9 × m!}{2!(m-2)!}~=~559$
$\Rightarrow~\frac{m!}{0!\;m!}~-~\frac{3m(m-1)!}{1!(m-1)!}~+~\frac{9m(m-1)(m-2)!}{2!(m-2)!}~=~559$
$\Rightarrow~1~-~\frac{3m}{1}~+~\frac{9m(m-1)}{2}~=~559$
$\Rightarrow~2~-~6m~+~9m(m-1)~=~559 × 2$
$\Rightarrow~2~-~6m~+~9m^2-9m~=~1118$
$\Rightarrow~9m^2 - 15m~=~1116$
• Solving this quadratic equation, we get: m = 12
5. Based on the result in (1), we can write:
For the power of x to be 3, [m-r] - 2r should be 3
So we get: [12-r] - 2r = 3
12 - 3r = 3
r = 3
6. Thus we can write:
In the term with r = 3, the power of x will be 3
• We get:
$T_{r+1}~=~T_{3+1}~=~{}^{12} {\rm{C}}_{3}~(-1)^3~\left(x \right)^{12-3}~\left(\frac{3}{x^2} \right)^r~=~-5940x^3$
Solved example 8.17
If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 + x)34 are equal, find r.
Solution:
1. We know that, the (r+1)th term of the binomial expansion (a+b)n is given by:
nCr an-r br
• In our present case, a = 1, b = x and n = 34.
• So we get:
(r+1)th term of the given binomial expansion is:
$T_{r+1}~=~{}^{34} {\rm{C}}_{r}~1^{34-r}~x^r~=~{}^{34} {\rm{C}}_{r}~x^r$
2. Based on this, we can write:
rth term of the given binomial expansion is:
$T_{r}~=~{}^{34} {\rm{C}}_{r-1}~x^{r-1}$
3. So the (r – 5)th term will be:
$T_{r-5}~=~{}^{34} {\rm{C}}_{(r-5)-1}~x^{(r-5)-1}~=~{}^{34} {\rm{C}}_{r-6}~x^{r-6}$
4. Similarly, the (2r – 1)th term will be:
$T_{2r-1}~=~{}^{34} {\rm{C}}_{(2r-1)-1}~x^{(2r-1)-1}~=~{}^{34} {\rm{C}}_{2r-2}~x^{2r-2}$
5. Given that, the coefficients are equal. So we can write:
${}^{34} {\rm{C}}_{r-6}~=~{}^{34} {\rm{C}}_{2r-2}$
6. Now we apply a property that we saw in the previous chapter:
If ${}^{n} {\rm{C}}_{a}~=~{}^{n} {\rm{C}}_{b}$, then a = b or n = a+b
[See result 6 in section 7.6]
• We can write:
(i) r - 6 = 2r-2 OR
(ii) r - 6 + 2r - 2 = 34
♦ From (i), we get: r = -4
♦ From (ii), we get: r = 14
• r must be a +ve integer. So r = 14
The link below gives some more miscellaneous examples.
Miscellaneous Exercise on chapter 8
In the next chapter we will see sequences and series.
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