# 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.

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