Chapter 8.5 - Miscellaneous Examples on Chapter 8

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:
ⁿC_r (-1)^r a^n-r b^r
• 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 a^r-1, a^r and a^r+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:
ⁿC_r a^n-r b^r
• 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: ⁿC_r 1^n-r a^r
= ⁿC_r a^r
3. It follows that:
• r^th term of the expansion of (1 + a)ⁿ will be: ⁿC_r-1 a^r-1
• (r+2)^th term of the expansion of (1 + a)ⁿ will be: ⁿC_r+1 a^r+1
4. So the three coefficients in arithmetic progression are:
ⁿC_r-1, ⁿC_r, ⁿC_r+1
• Since they are in arithmetic progression, the common difference will be the same. We can write:
ⁿC_r  - ⁿC_r-1 = ⁿC_r+1  - ⁿC_r 
^{⇒ n}C_r-1 + ⁿC_r+1 = 2 × ⁿC_r
• 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:
ⁿC_r a^n-r b^r
• 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: ²ⁿC_r 1^2n-r a^r
= ²ⁿC_r a^r
• Based on this, we can write:
Middle term, which is the (n+1)^th term will be: ²ⁿC_n aⁿ
• Coefficient of this middle term is ²ⁿC_n
• 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:
ⁿC_r a^n-r b^r
• 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-1C_r 1^2n-1-r a^r
= ^2n-1C_r a^r
• Based on this, we can write:
   ♦ The second middle term, which is the (n+1)^th term will be: ^2n-1C_n aⁿ
   ♦ The first middle term, which is the n^th term will be: ^2n-1C_n-1 a^n-1
• Sum of the coefficients of these two middle terms = ^2n-1C_n + ^2n-1C_n-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 r^th 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:
ⁿC_r a^n-r b^r
• In our present case, a = x and b = a.
• So we get:
(r+1)^th term of the binomial expansion (x+a)ⁿ is: ⁿC_r x^n-r a^r
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 ⁿC_n x^n-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 ⁿC_n x^n-n aⁿ 
4. Now, the n^th 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 n^th term will be ⁿC_n-1 x^n-(n-1) a^n-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 ⁿC_n-2 x^n-(n-2) a^n-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 ⁿC_n-3 x^n-(n-3) a^n-3 
7. Thus we see a pattern. Based on that pattern, we can write:
The r^th term from the end will be ⁿC_n-(r-1) x^n-[n-(r-1)] a^n-(r-1) 
• Simplifying this, we get: ⁿC_n-r+1 x^n-n+r-1 a^n-r+1 
= ⁿC_n-r+1 x^r-1 a^n-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:
ⁿC_r a^n-r b^r
• 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 10^th 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:
ⁿC_r (-1)^r a^n-r b^r
• 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:
ⁿC_r a^n-r b^r
• 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:
r^th 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

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The link below gives some more miscellaneous examples.

Miscellaneous Exercise on chapter 8

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In the next chapter we will see sequences and series.

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