Question

What is the Binomial Expansion of ${\left( {1 + r} \right)^{ - 1}}$ ?

Answer 1

A combination is the number of ways we can combine things, when the order does not matter.
Here we will apply the Binomial Expansion to solve the given problem.
The binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial ${(x + y)^n}$.
Formula Used:
Combination rule: $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Binomial expansion:
${(x + y)^n}{ = ^n}{C_0}{x^n}{ + ^n}{C_1}{x^{n - 1}}{y^1}{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......{ + ^n}{C_{n - 1}}x{y^{n - 1}}{ + ^n}{C_n}{y^n}$
Where $n \geqslant 0$, is an integer and each $^n{C_k}$ is a positive integer known as binomial coefficient.

Complete step-by-step solution:
We need to find out the binomial expansion of ${\left( {1 + r} \right)^{ - 1}}$.
Now we know that, according to binomial theorem it is possible to expand any nonnegative power of $x + y$ into a sum of the form
${(x + y)^n}{ = ^n}{C_0}{x^n}{ + ^n}{C_1}{x^{n - 1}}{y^1}{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......{ + ^n}{C_{n - 1}}x{y^{n - 1}}{ + ^n}{C_n}{y^n}$
Where $n \geqslant 0$,is an integer and each$^n{C_k}$is a positive integer known as binomial coefficient.
Now if we generalise the formula, for any arbitrary number m, one can define
$^m{C_k} = \dfrac{{m!}}{{(m - k)!k!}} = \dfrac{{m.\left( {m - 1} \right).\left( {m - 2} \right)......\left( {m - k + 1} \right)}}{{k!}}$
If $\left| x \right| > \left| y \right|$we can write, ${(x + y)^m} = \sum\limits_{k = 0}^\infty {^m{C_k}{x^{m - k}}{y^k}} { = ^m}{C_0}{x^m}{ + ^m}{C_1}{x^{m - 1}}{y^1}{ + ^m}{C_2}{x^{m - 2}}{y^2}{ + ^m}{C_3}{x^{m - 3}}{y^3} + ......$
$= {x^m} + m{x^{m - 1}}{y^1} + \dfrac{{m(m - 1)}}{{2!}}{x^{m - 2}}{y^2} + \dfrac{{m(m - 1)(m - 2)}}{{3!}}{x^{m - 3}}{y^3} + ......$
[Using the Combination rule,$^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$]
When m is a nonnegative integer, the binomial coefficients for $k > m$are zero, so this equation reduces to the usual binomial theorem, and there are at most $m + 1$ nonzero term. For other values of m, the series typically has infinitely many nonzero terms.
Now we can use the binomial expansion putting $x = 1,y = r\& m = - 1$we get,
${\left( {1 + r} \right)^{ - 1}} = 1 + ( - 1).1.r + \dfrac{{ - 1( - 1 - 1)}}{{2!}}.1.{r^2} + \dfrac{{ - 1( - 1 - 1)( - 1 - 2)}}{{3!}}.1.{r^3} + ......$
Solving we get,
${\left( {1 + r} \right)^{ - 1}} = 1 - r + \dfrac{{1.2}}{{2!}}{r^2} - \dfrac{{1.2.3}}{{3!}}{r^3} + ......$,
Therefore, we get,
${\left( {1 + r} \right)^{ - 1}} = 1 - r + {r^2} - {r^3} + ......$
[Using $n! = n(n - 1)(n - 2)(n - 4).......2.1$]
Hence expanding ${\left( {1 + r} \right)^{ - 1}}$ we get,
${\left( {1 + r} \right)^{ - 1}} = 1 - r + {r^2} - {r^3} + ......$

Note: In mathematics, a combination is a selection of items from a collection, such that the order of selection does not matter. For example, given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange.
For a combination,
$C\left( {n,r} \right)$=$^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Where, factorial n is denoted by n! and defined by
$n! = n(n - 1)(n - 2)(n - 4).......2.1$