Binomial Theorem For Rational Indices

Go back to  'Binomial Theorem'

In the previous section, we discussed the expansion of \({(x + y)^n}\) , where n is a natural number. We’ll extend that discussion to a more general scenario now. In particular, we’ll consider the expansion of \({(1 + x)^n}\) , where n is a rational number and | x | < 1. Note that any binomial of the form \({(a + b)^n}\) can be reduced to this form.:

\[\begin{align}{(a + b)^n} =& {\left\{ {a\left( {1 + \frac{b}{a}} \right)} \right\}^n} \qquad (we \;are \;assuming | a | > | b |)\\\\= &{a^n}{\left( {1 + \frac{b}{a}} \right)^n}\\\\= &{a^n}{(1 + x)^n}\qquad where\; | x | < 1\end{align}\]

The general binomial theorem states that

\[\fbox{${\begin{gathered} \;{(1 + x)^n} = 1 + nx + \frac{{n(n - 1)}}{{2!}}{x^2} + \frac{{n(n - 1)(n - 2)}}{{3!}}\;{x^3} + ........\; \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;........ + \frac{{n(n - 1)(n - 2)......(n - r + 1)}}{{r!}}{x^r} + .......\infty \; \\\end{gathered}}$} \]

That is, there are an infinite number of terms in the expansion with the general term given by

\[{T_{r + 1}} = \frac{{n(n - 1)(n - 2).......(n - r + 1)}}{{r!}}{x^r}\]

For an approximate proof of this expansion, we proceed as follows: assuming that the expansion contains an infinite number of terms, we have:

\[{\left( {1 + x} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ... + {a_n}{x^n} + ...\infty \]

Putting x = 0 gives a 0  = 1. Now differentiating once gives

\[n{\left( {1 + x} \right)^{n - 1}} = {a_1} + 2{a_2}x + 3{a_3}{x^2} + .....\infty \]

Putting x = 0 gives a 1  = n.

Proceeding in this way, we find that the r th  coefficient is given by

\[{a_n} = \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)...\left( {n - r + 1} \right)}}{{r!}}\]

Note that if n is a natural number, then this expansion reduces to the expansion obtained earlier, because \({T_{r + 1}}\) becomes \(^n{C_r}{x^r}\) , and the expansion terminates for \(r > n\) . For the general \({T_{r + 1}}\) , we obviously cannot use \(^n{C_r}\) .since that is defined only for natural n.

One very important point that we are emphasizing again is that the general expansion holds only for \(|x|\; < 1\) .

Let us denote the genral binomial coefficient by V r .  Thus, we have

\[{V_r} = \frac{{n(n - 1)(n - 2)......(n - r + 1)}}{{r!}}\]

and                 \[{(1 + x)^n} = \sum\limits_{r = 0}^\infty  {} {V_r}{x^r}\]

Let us discuss some particularly interesting expansions. In all cases \(|x|\; < \;1:\) ,

(1)   \({(1 + x)^{ - 1}}\) :                   Since   \(n =  - 1,\)   we see that

\[\begin{array}{l}{V_r} = \frac{{( - 1)(( - 1) - 1)(( - 1) - 2)......(( - 1) - r + 1)}}{{r!}}\\\,\,\,\,\, = {( - 1)^r}\end{array}\]

so that the expansion is

\[{(1 + x)^{ - 1}} = 1 - x + {x^2} - {x^3} + ......\infty \]

(2)   \({(1 - x)^{ - 1}}\) :           Again, \({V_r} = {( - 1)^r}\) and thus

\[{(1 - x)^{ - 1}} = 1 + x + {x^2} + {x^3} + ......\infty \]

(3) \({(1 + x)^{ - 2}}\) : We have n = – 2;

\[\begin{align}{V_r} =& \frac{{( - 2)(( - 2) - 1)(( - 2) - 2)....(( - 2) - r + 1)}}{{r!}}\\\\=&\frac{{{{( - 1)}^r}\;(r + 1)!}}{{r!}}\\\\=& {( - 1)^r} \cdot \;(r + 1)\end{align}\]

Thus,

\[{(1 + x)^{ - 2}} = 1 - 2x + 3{x^2} - 4{x^3} + ......\infty \]\\

(4) \({(1 - x)^{ - 2}}\) :                                                                     Again, \({V_r} = {( - 1)^r} \cdot (r + 1)\) so that

\[{(1 - x)^{ - 2}} = 1 + 2x + 3{x^2} + 4{x^3} + ........\infty \]

Example – 11

(a)   For \(|x|\; < \;1\) , expand \({(1 - x)^{ - 3}}\)

(b)    Find the coefficient of \({x^n}\) in the expansion of \({(1 + 3x + 6{x^2} + 10{x^3} + .......\infty )^{ - n}}\)

Solution: (a) We have

\[\begin{align}{}{V_r} = &\frac{{( - 3)\;(( - 3) - 1)(( - 3) - 2))......(( - 3) - r + 1)}}{{r!}}\\\,\,\,\,\, =& \frac{{{{( - 1)}^r}\;(r + 2)!}}{{2(r!)}}\\\,\,\,\,\, =& \frac{{{{( - 1)}^r}\;(r + 1)\;(r + 2)}}{2}\end{align}\]

Thus,

\[{V_0} = 1,\;\;{V_1} =  - 3,\;{V_2} = 6,\;{V_3} =  - 10........\]

so that

\[{(1 - x)^{ - 3}} = 1 + 3x + 6{x^2} + 10{x^3} + .......\infty \]

(b)            We use the result of part –(a) in this:

\[\begin{array}{l}{(1 + 3x + 6{x^2} + 10{x^3}.......\infty )^{ - n}} &= {\left( {{{(1 - x)}^{ - 3}}} \right)^{ - n}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= {(1 - x)^{3n}}\end{array}\]

The coefficient of \({x^n}\) in this binomial expansion (note: the power is now a positive integer) would be \({( - 1)^n} \cdot {\;^{3n}}{C_n}\) .

Example – 12

Find the magnitude of the greatest term in the expansion of \({\left( {1 - 5y} \right)^{ - 2/7}}\) for \(y = \begin{align}\frac{1}{8}\end{align}\) .

Solution:    Let us first do the general case: what is the greatest term in the expansion of \({(1 + x)^n}\) , where n is  an arbitrary rational number. We have,

\[\begin{align}{T_{r + 1}} &= {V_r}\;{x^r}\\\\And\quad {T_r} &= {V_{r - 1}}\;{x^{r - 1}}\\\\so \;\;that\;\; \frac{{{T_{r + 1}}}}{{{T_r}}} &= \frac{{{V_r}}}{{{V_{r - 1}}}} \cdot x\\\\&= \frac{{n - r + 1}}{r} \cdot x\end{align}\]

Now, let us find the conditions for which this ratio exceeds 1. We have

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\left| {\;{T_{r + 1}}\;} \right|\; &\ge \;\left| {\;{T_r}\;} \right|\\\\ \Rightarrow \,\,\,\,\,\,\,\left| {\;\frac{{n + 1}}{r} - 1} \right|\; &\ge \;\frac{1}{{|x|}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\end{align}\]

For this particular problem, (1) becomes

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\left| {\;\frac{{\frac{{ - 2}}{7} + 1}}{r} - 1} \right|\; \ge \;\frac{1}{{\left| {\frac{{ - 5}}{8}} \right|}}\\\\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,&\left| {\;\frac{5}{{7r}} - 1\;} \right|\; &\ge \;\frac{8}{5}\\\\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,&\frac{5}{{7r}}\; \ge \;\frac{{13}}{5}\\\\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,&r \le \;\frac{{25}}{{91}}\\\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,&r = 0\end{align}\]

Thus, \({T_{r + 1}} = {T_1}\) is the greatest term, with magnitude 1.

Download practice questions along with solutions for FREE:
Binomial Theorem
grade 11 | Questions Set 1
Binomial Theorem
grade 11 | Answers Set 1
Binomial Theorem
grade 11 | Questions Set 2
Binomial Theorem
grade 11 | Answers Set 2
Download practice questions along with solutions for FREE:
Binomial Theorem
grade 11 | Questions Set 1
Binomial Theorem
grade 11 | Answers Set 1
Binomial Theorem
grade 11 | Questions Set 2
Binomial Theorem
grade 11 | Answers Set 2
Learn math from the experts and clarify doubts instantly

  • Instant doubt clearing (live one on one)
  • Learn from India’s best math teachers
  • Completely personalized curriculum