A.2 Exponential Series

In the previous section, we saw infinite geometric series. In this section, we will see exponential series.

• Consider the series:
$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} +~.~.~$
• The sum of this series is denoted by the letter e.
• e is very important for calculus and other areas of mathematics, economics, financial studies, science and engineering.
• Note that all terms in the series, are +ve. So e will be +ve.

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Let us find the value of e. It can be done in 8 steps:

1. Two series are written below. They are written in such a way that, the terms in the second series come directly below the corresponding terms in the first series.

\begin{array}{ll}{}    &{\frac{1}{3!}}    & {~+~}    &{\frac{1}{4!}}    & {~+~}    &{\frac{1}{5!}}    & {~+~.~.~.~+~}    &{\frac{1}{n!}}    & {~+~.~.~.~\color{green}{\text{- - - (I)}}}    &{} \\
{}    &{\frac{1}{2^2}}    & {~+~}    &{\frac{1}{2^3}}    & {~+~}    &{\frac{1}{2^4}}    & {~+~.~.~.~+~}    &{\frac{1}{2^{n-1}}}    & {~+~.~.~.~\color{green}{\text{- - - (II)}}}    &{} \\
\end{array}

2. We see an interesting point. It can be written in 3 steps:
(i) Take any term in (II). Note down the power of 2 in that term.
(ii) Take the corresponding term in (I). Note down the denominator of that term.
(iii) The power will be one less than the denominator.

3. Another interesting point can also be observed. It can be written in 4 steps:
(i) Take the first terms: $\frac{1}{3!}~\text{and}~\frac{1}{2^2}$
• We have:
$\frac{1}{3!} = \frac{1}{6}~\text{and}~\frac{1}{2^2} = \frac{1}{4}$
• So the first term in (I) is less than the corresponding term in (II)

(ii) Take the second terms: $\frac{1}{4!}~\text{and}~\frac{1}{2^3}$
• We have:
$\frac{1}{4!} = \frac{1}{24}~\text{and}~\frac{1}{2^3} = \frac{1}{8}$
• So the second term in (I) is less than the corresponding term in (II)

(iii) Take the third terms: $\frac{1}{5!}~\text{and}~\frac{1}{2^4}$
• We have:
$\frac{1}{5!} = \frac{1}{102}~\text{and}~\frac{1}{2^4} = \frac{1}{16}$
• So the third term in (I) is less than the corresponding term in (II)

(iv) Based on the above 3 steps, we can write:
Whenever n is greater than 2,
$\frac{1}{n!}~<~\frac{1}{2^{n-1}}$

4. We saw that, each term in (I) is less than the corresponding term in (II). So we can write:

$\begin{array}{ll}{}    &{\frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.}    & {~<~}    &{\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.}    &{} \\
    {\Rightarrow}    &{\left(1 + \frac{1}{1!} + \frac{1}{2!} \right)~+~\left(\frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.\right)}    & {~<~}    &{\left(1 + \frac{1}{1!} + \frac{1}{2!} \right)~+~\left(\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.\right)~\color{green}{\text{- - - (A)}}}    &{} \\
    {\Rightarrow}    &{\left(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.\right)}    & {~<~}    &{\left(1 + 1 + \frac{1}{2^1} \right)~+~\left(\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.\right)}    &{} \\
    {\Rightarrow}    &{1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.}    & {~<~}    &{1 + \left(1 + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.\right)~\color{green}{\text{- - - (B)}}}    &{} \\
    {\Rightarrow}    &{1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.}    & {~<~}    &{1 + \frac{1}{1 - \frac{1}{2}}}    &{} \\
    {\Rightarrow}    &{1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.}    & {~<~}    &{3}    &{} \\
    {\Rightarrow}    &{e}    & {~<~}    &{3}    &{} \\
    \end{array}$

◼ Remarks:
• Line marked as A:
Here we add $\left(1 + \frac{1}{1!} + \frac{1}{2!} \right)$ on both sides.
• Line marked as B:
Consider the RHS of this line. We have a portion enclosed in brackets. This portion is a infinite geometric series. It’s sum can be easily calculated. It works out to 2.

5. Earlier we said that:
$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} +~.~.~=~e$
• Consider the LHS. The first two terms are 1 and 1.
• So we can write:
e will be always greater than 2.

6. Let us compare the results:
    ♦ From (4) we have: e < 3
    ♦ From (5) we have: e > 2
• So we can write: 2 < e < 3
7. The approximate value of e is 2.71828. The accuracy will increase when we increase the number of terms in the series.
• e is an irrational number like π.
8. Consider the equation that we use to evaluate e:
$e~=~1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} +~.~.~.$
    ♦ In the LHS, the power of e is 1.
    ♦ In the RHS, all numerators are 1.
        ✰ These ones are actually, powers of 1.
• So we can write:
$e^1~=~1 + \frac{1^1}{1!} + \frac{1^2}{2!} + \frac{1^3}{3!} + \frac{1^4}{4!} +~.~.~.$
• If the power of e is 2, then we can write:
$e^2~=~1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} +~.~.~.$     
• If the power of e is 7, then we can write:
$e^7~=~1 + \frac{7^1}{1!} + \frac{7^2}{2!} + \frac{7^3}{3!} + \frac{7^4}{4!} +~.~.~.$
◼ We can write the general form:
• If the power of e is x, then:
$e^x~=~1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} +~.~.~.~+ \frac{x^n}{n!}~+~.~.~.$

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Now we will see a solved example:

Solved example A.3
Find the coefficient of x² in the expansion of e^2x+3 as a series in the powers of x.
Solution:
1. We have the general form:
$e^x~=~1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} +~.~.~.~+ \frac{x^n}{n!}~+~.~.~.$
2. In our present case, we have (2x+3) in the place of x. So we can write:
$e^{2x+3}~=~1 + \frac{(2x+3)^1}{1!} + \frac{(2x+3)^2}{2!} + \frac{(2x+3)^3}{3!} + \frac{(2x+3)^4}{4!} +~.~.~.~+ \frac{(2x+3)^n}{n!}~+~.~.~.$
3. The general term is: $\frac{(2x+3)^n}{n!}$
The numerator of this general term can be expanded using binomial theorem. We get:

$\begin{array}{ll}{}    &{\frac{(2x+3)^n}{n!}}    & {~=~}    &{\frac{(3+2x)^n}{n!}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1}{n!}\left[3^n + \rm{{}^n C_1}3^{n-1}(2x) +\rm{{}^n C_2}3^{n-2}(2x)^2 + \rm{{}^n C_3}3^{n-3}(2x)^3~+~.~.~.~+~(2x)^n  \right]}    &{} \\
\end{array}$

4. Note the third term of the above expansion. It is the term with "x²".
• So we can write:
General form of the coefficient of x² is $\frac{\rm{{}^n C_2}3^{n-2}(2x)^2}{n!}$

5. In (3), we considered the general term. For each term, there will be a term with x². We want the sum of coefficients of all such terms. That sum will be:
$\sum_{n=2}^{n=\infty}{\frac{\rm{{}^n C_2}3^{n-2}2^2}{n!}}$
(We saw similar problems in section 9.5) 

• This can be simplified as shown below:

$\begin{array}{ll}{}    &{\sum_{n=2}^{n=\infty}{\frac{\rm{{}^n C_2}3^{n-2}2^2}{n!}}}    & {~=~}    &{\sum_{n=2}^{n=\infty}{\frac{\frac{n!}{2! \times (n-2)!}\times 3^{n-2} \times 4}{n!}}}    &{} \\
{}    &{}    & {~=~}    &{\sum_{n=2}^{n=\infty}{\frac{\frac{1}{1 \times (n-2)!}\times 3^{n-2} \times 2}{1}}}    &{} \\
{}    &{}    & {~=~}    &{2\sum_{n=2}^{n=\infty}{\frac{3^{n-2}}{(n-2)!}}}    &{} \\
{}    &{}    & {~=~}    &{2 \left[\frac{3^{2-2}}{(2-2)!} + \frac{3^{3-2}}{(3-2)!} + \frac{3^{4-2}}{(4-2)!} + \frac{3^{5-2}}{(5-2)!}~+~.~.~. \right]}    &{} \\
{}    &{}    & {~=~}    &{2 \left[\frac{3^{0}}{0!} + \frac{3^{1}}{1!} + \frac{3^{2}}{2!} + \frac{3^{3}}{3!}~+~.~.~. \right]}    &{} \\
{}    &{}    & {~=~}    &{2 \left[\frac{1}{1} + \frac{3^{1}}{1!} + \frac{3^{2}}{2!} + \frac{3^{3}}{3!}~+~.~.~. \right]}    &{} \\
{}    &{}    & {~=~}    &{2 \left[1 + \frac{3^{1}}{1!} + \frac{3^{2}}{2!} + \frac{3^{3}}{3!}~+~.~.~. \right]}    &{} \\
{}    &{}    & {~=~}    &{2e^3}    &{} \\
\end{array}$

6. So the coefficient of x² is 2e³.

Alternate method:

1. $\begin{array}{ll}{}    &{e^{2x+3}}    & {~=~}    &{e^3 \times e^{2x}}    &{} \\
{}    &{}    & {~=~}    &{e^3 \left[1 + \frac{2x}{1!} + \frac{(2x)^{2}}{2!} + \frac{(2x)^{3}}{3!}~+~.~.~. \right]}    &{} \\
\end{array}$
2. Consider the third term in the above result. It is the term with x².
• This term can be written as: $\frac{4x^2 e^3}{2!}~=~2x^2 e^3$.
3. So we can write:
The coefficient of x² is $2e^3$

Solved example A.4
Find the value of e² rounded off to one decimal place.
Solution:
1. Let us expand e²:
$\begin{array}{ll}{}    &{e^2}    & {~=~}    &{1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} \right) + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + \frac{8}{6} + \frac{16}{24} \right) + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)}    &{} \\
{}    &{}    & {~=~}    &{\left(5 + \frac{4}{3} + \frac{2}{3} \right) + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)}    &{} \\
{}    &{}    & {~=~}    &{7 + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)}    &{} \\
\end{array}$

2. In the last line of the above result, consider the portion inside brackets. We will modify that portion as shown below:
$\begin{array}{ll}{}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~=~}    &{\frac{2^5}{5!} + \frac{2^6}{6 \times 5!} + \frac{2^7}{7 \times 6 \times 5!} + \frac{2^8}{8 \times 7 \times 6 \times 5!}~+~.~.~.}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~=~}    &{\frac{1}{5!}\left(\frac{2^5}{1} + \frac{2^6}{6} + \frac{2^7}{7 \times 6} + \frac{2^8}{8 \times 7 \times 6 }~+~.~.~. \right)}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{1}{5!}\left(\frac{2^5}{1} + \frac{2^6}{6} + \frac{2^7}{6 \times 6} + \frac{2^8}{6 \times 6 \times 6 }~+~.~.~. \right)~\color{green}{{\text{- - - (A)}}}}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{1}{5!}\left(\frac{2^5}{1-\frac{2}{6}} \right)~\color{green}{\text{- - - (B)}}}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{1}{5!}\left(\frac{2^5}{\frac{2}{3}} \right)}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{1}{5 \times 4 \times 3 \times 2 \times 1}\left(\frac{2^5 \times 3}{2} \right)}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{1}{5 \times 1 \times 1 \times 1 \times 1}\left(\frac{2 \times 1}{1} \right)}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{2}{5}}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{0.4}    &{} \\
\end{array}$

◼ Remarks:
• Line marked as (A):
In this line, consider the denominators of the RHS.
   ♦ We put 6 in the place of 7.
   ♦ We put 6 in the place of 8 also.
         ✰ So the denominators decrease. As a result, the fractions increase in values.
         ✰ Then the whole RHS becomes larger than the LHS. We change the "=" sign to "<" sign.
• Line marked as (B):
In the line line marked as (A), in the RHS, the portion inside brackets is an infinite geometric series. It's sum can be easily calculated.

3. Now consider the last line of the result in (1).
Imagine that we replace $\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.$
by $\frac{1}{5!}\left(\frac{2^5}{1} + \frac{2^6}{6} + \frac{2^7}{6 \times 6} + \frac{2^8}{6 \times 6 \times 6 }~+~.~.~. \right)$
• Then we can write: e² < 7.4

4. Consider again the expansion of e². We want the first seven terms:

$\begin{array}{ll}{}    &{e^2}    & {~=~}    &{1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} + \frac{2^6}{6!} \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + \frac{2^3}{3 \times 2 \times 1} + \frac{2^4}{4 \times 3 \times 2 \times 1} + \frac{2^5}{5 \times 4 \times 3 \times 2 \times 1} + \frac{2^6}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + \frac{4}{3} + \frac{2}{3} + \frac{4}{15} + \frac{4}{45} \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + \frac{60 + 30 + 12 + 4}{45} \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + \frac{106}{45} \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + 2.3555556 \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(7.355 \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
\end{array}$

• It is clear that, e² is greater than 7.355556.
• This value when rounded off to one decimal place is: 7.4.

5. Let us compare the results:
   ♦ From (3), we have: e² < 7.4
   ♦ From (4), we have: e² > 7.4

6. So the value of e² rounded off to one decimal place is 7.4.

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In the next section, we will see mathematical modelling.

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Appendix A - Infinite Series

In the previous section, we completed a discussion on probability. In this appendix A, we will see Infinite series.

• In chapter 9, we discussed sequences and series.
1. Consider the sequence given below:
$a_1, a_2, a_3,~.~.~.~,~a_n,~.~.~.$
• This sequence has infinite number of terms. So it is called an infinite sequence.
2. Let us write the corresponding series:
$a_1~+~a_2~+~a_3~+~.~.~.~,~+~a_n~+~.~.~.$
• This series is called the infinite series associated with the infinite sequence.
3. We can abbreviate an infinite series as shown below:
$a_1~+~a_2~+~a_3~+~.~.~.~,~+~a_n~+~.~.~.~=~\sum\limits_{i=1}^{i=\infty} a_i$

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• In chapter 8, we discussed binomial theorem. Based on that theorem, we derived the formula:
$(1+x)^n~=~{}^n \rm{C}_0~+~{}^n \rm{C}_1 x~+~{}^n \rm{C}_2 x^2~+~{}^n \rm{C}_3 x^3~+~.~.~.~+~{}^n \rm{C}_n x^n$
• "n" in the above formula must be an integer. Also, it should not be -ve. This is because, in such cases, we will not be able to calculate ${}^n \rm{C}_r$.
• We will now see a formula which can be used when n is not an integer and/or n is -ve. It can be explained in 4 steps:
1. The formula is:
$(1+x)^m~=~1~+~mx~+~{\frac{m(m-1)}{1 \times 2}}x^2~+~{\frac{m(m-1)(m-2)}{1 \times 2 \times 3}}x^3~+~{\frac{m(m-1)(m-2)(m-3)}{1 \times 2 \times 3 \times 4}}x^4~+~.~.~.$
• We will see the derivation of this formula in higher classes.
2. This formula is applicable whenever |x| < 1
• This condition can be explained in 5 steps:
(i) Consider the value of x. Based on that value, we can mark it’s position on the number line.
(ii) The distance of this mark from zero must be less than 1.
(iii) So, if x is -ve,
    ♦ The mark must not be on -1.
    ♦ The mark must not be anywhere to the left of -1.
(iv) Similarly, if x is +ve,
    ♦ The mark must not be on 1.
    ♦ The mark must not be anywhere to the right of 1.
(v) We can combine the above four points as:
-1 < x < 1

• The importance of this condition will become clear when we see an example. Let us put “-3” in the place of x and "-2" in the place of m. We get:

$\begin{array}{ll}{}    &{(1-3)^{-2}}    & {~=~}    & {1 + (-2)(-3) + \frac{(-2)(-3-1)}{1 \times 2} (-3)^2 + \frac{(-2)(-3-1)(-3-2)}{1 \times 2 \times 3} (-3)^3 + . . .}    &{} \\
{\Rightarrow}    &{(-2)^{-2}}    & {~=~}    & {1 + 6 + \frac{(-2)(-4)}{2} (9) + \frac{(-2)(-4)(-5)}{6} (-27) + . . .}    &{} \\
{\Rightarrow}    &{(-1)^{-2} \times (2)^{-2}}    & {~=~}    & {1 + 6 + \frac{8}{2} (9) + \frac{-40}{6} (-27) + . . .}    &{} \\
{\Rightarrow}    &{1 \times \frac{1}{4}}    & {~=~}    & {1 + 6 + (4) (9) + \frac{-20}{3} (-27) + . . .}    &{} \\
{\Rightarrow}    &{\frac{1}{4}}    & {~=~}    & {1 + 6 + 36 + 180 + . . .}    &{} \\
\end{array}$

• This is not possible. So before applying this formula, we must make sure that -1 < x < 1.

3. We know how to expand (a+b)^m when m is a +ve integer. But what if m is not an integer and/or -ve?
In such cases, we can use the above formula. This is shown below:

$\begin{array}{ll}{}    &{(a+b)^m}    & {~=~}    & {\left[a \left(1 + \frac{b}{a} \right) \right]^m}    &{} \\
{}    &{}    & {~=~}    & {a^m \left(1 + \frac{b}{a} \right)^m}    &{} \\
{}    &{}    & {~=~}    & {a^m \left[1 + m{\frac{b}{a}} + {\frac{m(m-1)}{1 \times 2}}\left(\frac{b}{a} \right)^2 + {\frac{m(m-1)(m-2)}{1 \times 2 \times 3}}\left(\frac{b}{a} \right)^3~+~.~.~. \right]}    &{} \\
{}    &{}    & {~=~}    & {a^m + m a^{m-1} b + {\frac{m(m-1)}{1 \times 2}}a^{m-2} b^2 + {\frac{m(m-1)(m-2)}{1 \times 2 \times 3}}a^{m-3} b^3 ~+~.~.~.}    &{} \\
\end{array}$

• We were able to use the formula because we wrote (a+b)^m as $\left[a \left(1 + \frac{b}{a} \right) \right]^m$.
• Recall that, to use the formula, |x| must be less than 1.
• So in our present case, $\left| \frac{b}{a} \right| $ must be less than 1.
• This can be explained in 5 steps:
(i) Values of a and b:
    ♦ a can be +ve or -ve. It can be any real number.
    ♦ b can be +ve or -ve. It can be any real number.
(ii) Calculating $\frac{b}{a}$:
We calculate $\frac{b}{a}$ using the proper signs.
(iii) Marking the position on the number line:
We mark $\frac{b}{a}$ based on the result in (ii).
(iv) The mark can be either on the left side or right side of zero. But it's distance from zero must be less than 1.

Let us see some examples:
Example1:
• Put a = 2 and b = -3
• Then $\frac{b}{a} = \frac{-3}{2}$ = -1.5
• When we mark $\frac{b}{a}$ on the number line, it will be at a distance of 1.5 units from zero. So we will not be able to use the formula.

Example2:
• Put a = 4 and b = -3
• Then $\frac{b}{a} = \frac{-3}{4}$ = -0.75
• When we mark $\frac{b}{a}$ on the number line, it will be at a distance of 0.75 units from zero. So we can use the formula

(v) Based on the above four steps, we can write:
$\frac{b}{a}$ must be a proper fraction. It can be +ve or -ve.

4. Based on the expansion in (3), we can write the general term in the expansion. It is given below:

$$\frac{m(m-1)(m-2)(m-3)~.~.~.~(m-r+1) a^{m-r} b^r}{1 \times 2 \times 3 \times ~.~.~.\times r}$$

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Based on the formula that we wrote in (1), we can derive four useful results:

Result I:
$\begin{array}{ll}{}    &{(1+x)^{-1}}    & {~=~}    &{1 + (-1)x + \frac{(-1)(-1-1)}{1 \times 2} x^2 + \frac{(-1)(-1-1)(-1-2)}{1 \times 2 \times 3} x^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + (-1)x + \frac{(-1)(-2)}{1 \times 2} x^2 + \frac{(-1)(-2)(-3)}{1 \times 2 \times 3} x^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + (-1)x + \frac{(-1)(-1)}{1 \times 1} x^2 + \frac{(-1)(-1)(-1)}{1 \times 1 \times 1} x^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 - x + x^2 – x^3~+~.~.~.}    &{} \\
\end{array}$

Result II:                

$\begin{array}{ll}{}    &{(1-x)^{-1}}    & {~=~}    &{[1+(-x)]^{-1}}    &{} \\
{}    &{}    & {~=~}    &{1 + (-1)(-x) + \frac{(-1)(-1-1)}{1 \times 2} (-x)^2 + \frac{(-1)(-1-1)(-1-2)}{1 \times 2 \times 3} (-x)^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + (-1)(-x) + \frac{(-1)(-2)}{1 \times 2} (x^2) + \frac{(-1)(-2)(-3)}{1 \times 2 \times 3} (-1)(x^3)~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + (-1)(-x) + \frac{(-1)(-1)}{1 \times 1} (x^2) + \frac{(-1)(-1)(-1)}{1 \times 1 \times 1} (-1)(x^3)~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + x + x^2 + x^3~+~.~.~.}    &{} \\
\end{array}$

Result III:

$\begin{array}{ll}{}    &{(1+x)^{-2}}    & {~=~}    &{1 + (-2)x + \frac{(-2)(-2-1)}{1 \times 2} x^2 + \frac{(-2)(-2-1)(-2-2)}{1 \times 2 \times 3} x^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + (-2)x + \frac{(-2)(-3)}{1 \times 2} x^2 + \frac{(-2)(-3)(-4)}{1 \times 2 \times 3} x^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + (-2)x + \frac{(-1)(-3)}{1 \times 1} x^2 + \frac{(-1)(-1)(-4)}{1 \times 1 \times 1} x^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 - 2x + 3 x^2 –4 x^3~+~.~.~.}    &{} \\
\end{array}$

Result IV:

$\begin{array}{ll}{}    &{(1-x)^{-2}}    & {~=~}    &{[1+(-x)]^{-2}}    &{} \\
{}    &{}    & {~=~}    &{1 + (-2)(-x) + \frac{(-2)(-2-1)}{1 \times 2} (-x)^2 + \frac{(-2)(-2-1)(-2-2)}{1 \times 2 \times 3} (-x)^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + (-2)(-x) + \frac{(-2)(-3)}{1 \times 2} (-x)^2 + \frac{(-2)(-3)(-4)}{1 \times 2 \times 3} (-x)^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + (-2)(-x) + \frac{(-1)(-3)}{1 \times 1} x^2 + \frac{(-1)(-1)(-4)}{1 \times 1 \times 1} (-1) x^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + 2x + 3x^2 + 4x^3~+~.~.~.}    &{} \\
\end{array}$

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Now we will see a solved example:

Solved example 1
Expand $\left(1 - \frac{x}{2} \right)^{- \frac{1}{2}}$, when |x| < 2.
Solution:
1. We can rewrite the given expression as:
$\left[1 + \left(- \frac{x}{2} \right) \right]^{- \frac{1}{2}}$
• Consider the term $\left(- \frac{x}{2} \right)$.
We know that, this term must be less than 1 and at the same time, greater than -1. This condition will be satisfied only if |x| is less than 2. The reader must do the necessary analysis and become convinced about this fact.
2. Now we can write the expansion:

$\begin{array}{ll}{}    &{\left[1 + \left(- \frac{x}{2} \right) \right]^{- \frac{1}{2}}}    & {~=~}    &{1 + \left(-\frac{1}{2} \right) \left(-\frac{x}{2} \right) + \frac{\left(-\frac{1}{2} \right)\left(-\frac{1}{2} – 1 \right)}{1 \times 2} \left(-\frac{x}{2} \right)^2 + \frac{\left(-\frac{1}{2} \right)\left(-\frac{1}{2} – 1 \right)\left(-\frac{1}{2} – 2 \right)}{1 \times 2 \times 3} \left(-\frac{x}{2} \right)^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + \left(-\frac{1}{2} \right) \left(-\frac{x}{2} \right) + \frac{\left(-\frac{1}{2} \right)\left(-\frac{3}{2} \right)}{1 \times 2} \left(-\frac{x}{2} \right)^2 + \frac{\left(-\frac{1}{2} \right)\left(-\frac{3}{2} \right)\left(-\frac{5}{2} \right)}{1 \times 2 \times 3} \left(-\frac{x}{2} \right)^3~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + \left(-\frac{1}{2} \right) \left(-\frac{x}{2} \right) + \frac{3}{8} \left(-\frac{x}{2} \right)^2 + \frac{-5}{16} \left(-\frac{x}{2} \right)^3 ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{1 + \frac{x}{4} +\frac{3 x^2}{32} + \frac{5 x^3}{128} ~+~.~.~.}    &{} \\
\end{array}$

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In the next section, we will see infinite geometric series.

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