Examples On Binomial Theorem For Positive Integer Indices

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Example – 3

Evaluate the sum \(^n{C_0} + {\,^n}{C_1} + {\,^n}{C_2} + ... + {\,^n}{C_n}\) .

Solution:    We have already evaluated this sum in the chapter on P & C.  That approach was as follows: this sum basically counts the number of all sub-groups of a set of size n ; this can also be counted by focusing on each element of the set, which has two corresponding choices - you either include it into your sub-group or you don’t, which means that the total number of ways to form sub-groups is 2 × 2 × 2 .... n times = 2 n . The sum of the binomial coefficients therefore equals 2 n .

Here, we evaluate the same sum using a binomial approach. Consider the following expansion:

\[{\left( {1 + x} \right)^n} = \,{\,^n}{C_0} + {\,^n}{C_1}x + {\,^n}{C_2}\,{x^2} + {\,^n}{C_3}\,{x^3} + ... + {\,^n}{C_n}\,{x^n}\]

If we put x = 1, we simply obtain

\({2^n} = {\,^n}{C_0} + {\,^n}{C_1} + {\,^n}{C_2} + ...{\,^n}{C_n}\)

Thus, the same result is obtainable from both a combinatorial and a binomial approach.

We can also derive another useful result by putting x = –1 in the above relation, so that we obtain

\[\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 = {\,^n}{C_0} - {\,^n}{C_1} + {\,^n}{C_2} - {\,^n}{C_3} + ... + {\left( { - 1} \right)^n}{\,^n}{C_n}\\ \Rightarrow  \qquad { ^n}{C_0} + {\,^n}{C_2} + {\,^n}{C_4} + ... = {\,^n}{C_1} + {\,^n}{C_3} + {\,^n}{C_5} + ...\end{array}\]

This states the sum of the even-numbered coefficients is equal to the sum of the odd-numbered coefficients. Can you prove this using a combinatorial approach?

As an exercise, prove the following relations:

\[\begin{array}{l}^n{C_0}{2^n} + {\,^n}{C_1}{2^{n - 1}} + {\,^n}{C_2}\,{2^{n - 2}} + ...{\,^n}{C_n} = {3^n} + \\\\^n{C_0} - {\,^n}{C_1}x + {\,^n}{C_2}\,{x^2}\,\, - ...\, + {\left( { - 1} \right)^n}{\,^n}{C_n}{x^n} = {\left( {1 - x} \right)^n}\\\\^n{C_0} - \,\frac{{^n{C_1}}}{2} + \,\frac{{^n{C_2}}}{{{2^2}}}\, - \frac{{^n{C_3}}}{{{2^3}}} + ...\, + \frac{{{{\left( { - 1} \right)}^n}{\,^n}{C_n}}}{{{2^n}}} = \frac{1}{{{2^n}}}\end{array}\]

Example – 4

(a)     What is the greatest coefficient in the expansion of \({\left( {x + y} \right)^n}\) ?

(b)     What is the greatest term in the expression of \({\left( {x + y} \right)^n}\) ?

Solution:   (a) For this part, we basically need to only determine \(\max \left( {^n{C_r}} \right){\rm{for}}\,\,0 \le r \le n\,;\) x and y have no role to play in this part.

To find the greatest coefficient, consider the following ratio:

\[\begin{array}{l}q = \frac{{^n{C_{r + 1}}}}{{^n{C_r}}} = \frac{{\frac{{n!}}{{\left( {r + 1} \right)!\left( {n - r - 1} \right)!}}}}{{\frac{{n!}}{{r!\left( {n - r} \right)!}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n - r}}{{r + 1}}\end{array}\]

Thus,

\[\begin{array}{l}q > 1 \qquad  \Rightarrow  \qquad \frac{{n - r}}{{r + 1}} > 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow  \qquad n - r > r + 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow  \qquad r < \frac{{n - 1}}{2}\end{array}\]

Similarly,

\[\begin{array}{l}q < 1 \qquad  \Rightarrow  \qquad \frac{{n - r}}{{r + 1}} < 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow  \qquad n - r < r + 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow  \qquad r > \frac{{n - 1}}{2}\end{array}\]

Thus,

\[\left. \begin{array}{l} { ^n}{C_{r + 1}} > {\,^n}{C_r} \qquad {\rm{whenever}} \qquad r < \frac{{n - 1}}{2}\\{\rm{and}} \qquad { ^n}{C_{r + 1}} < {\,^n}{C_r} \qquad {\rm{whenever}} \qquad r > \frac{{n - 1}}{2}\,\end{array} \right\}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\]

If n is odd, we have

\[^n{C_{\frac{{n - 1}}{2}}} > {\,^n}{C_{\frac{{n - 3}}{2}}}\]

and \[^n{C_{\frac{{n + 3}}{2}}} < {\,^n}{C_{\frac{{n + 1}}{2}}}\]

Also, since

\[^n{C_{\frac{{n - 1}}{2}}} = {\,^n}{C_{\frac{{n + 1}}{2}}}\]

we see that for odd n, the two middle coefficients are the greatest. This can be verified by considering the following expansion:

\[\begin{array}{l}{\left( {x + y} \right)^5} = {x^5} + 5{x^4}y + \mathop {10}\limits_ \nwarrow  {x^3}{y^2} + \mathop {10}\limits_ \nearrow  {x^2}{y^3} + 5x{y^4} + {y^5}\\ \qquad \,\,{ \qquad ^{\scriptstyle\,\,\,\,\,\,\,\,\,\,\,\,{\rm{The}}\,\,{\rm{two}}\,\,{\rm{middle coefficients are }}\atop\scriptstyle\,\,\,\,\,\,\,\,\,\,\,\,{\rm{the}}\,\,{\rm{greatest}}\,\,{\rm{for}}\,\,{\rm{odd}}\,\,\,n}}\,\, \end{array}\]

If n is even, (1) gives

\[^n{C_{\frac{n}{2}}} > {\,^n}{C_{\frac{n}{2} - 1}}\]

and   \(^n{C_{\frac{n}{2} + 1}} < \,{\,^n}{C_{\frac{n}{2}}}\)

In this case therefore, the greatest coefficient is the single middle coefficient \(^n{C_{\frac{n}{2}}}\) . Lets verify this again:

\[\begin{array}{l}{\left( {x + y} \right)^6} = {x^6} + 6{x^5}y + 15{x^4}{y^2} + \mathop {20}\limits_ \nearrow  \,{x^3}{y^3} + 15{x^2}{y^4} + 6x{y^5} + {y^6}\\\qquad \,\,{ ^{\scriptstyle\, \qquad \,\,\,{\rm{The}}\,\,{\rm{single}}\,\,{\rm{middle coefficient is }}\atop\scriptstyle\, \qquad \,\,\,{\rm{the}}\,\,{\rm{greatest}}\,\,{\rm{for}}\,\,{\rm{even}}\,\,\,n}}\,\, \qquad \end{array}\]

(b) To find the greatest term, we must also consider x and y.  We again follow the approach of part (a):

\[\begin{align}{}q = \frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{^n{C_r}{x^{n - r}}{y^r}}}{{^n{C_{r - 1}}{x^{n - r + 1}}{y^{r - 1}}}}\\\\\,\,\, = \frac{{\left( {n - r + 1} \right)}}{r}.\frac{y}{x}\end{align}\]

Observe that

\[\begin{align}{}q > 1 \qquad  \Rightarrow  \qquad \frac{{\left( {n - r + 1} \right)}}{r} \cdot \frac{y}{x} > 1\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow  \qquad \left\{ {\frac{{\left( {n + 1} \right)y}}{{x + y}} - r} \right\} \cdot \left( {\frac{{x + y}}{{r\,x}}} \right) > 0\end{align}\]

If \(\begin{align}\frac{{\left( {n + 1} \right)y}}{{x + y}}\end{align}\) is an integer m, which must lie in (0, n ], we see that there are two greatest terms T m  and  T m + 1 . (Why).  Here’s the explanation:

We have

\(q > 1\) or  \(1 \le r \le m\)   and \(q < 1\) for \(r > m\)

\( \Rightarrow  \qquad {T_r} < {T_{r + 1}}\)   for         \(1 \le r \le m\) and       \({T_r} > {T_{r + 1}}\)   for  \(r > m\)

\( \Rightarrow  \qquad {T_{m - 1}} < {T_m}\,\,,\,\,{T_{m + 1}} > {T_{m + 2}},\,\,{T_m} = {T_{m + 1}}\)

\( \Rightarrow  \qquad {T_m}\,\,{\rm{and}}\,\,{T_{m + 1}}\) are the two greatest terms

Now, if \(\begin{align}\frac{{\left( {n + 1} \right)y}}{{x + y}}\end{align}\) is a non-integer, assume \(\begin{align}\left[ {\frac{{\left( {n + 1} \right)y}}{{x + y}}} \right] = m\end{align}\) .

We now have

\(q > 1\)   for  \(1 \le r \le m\) and \(q < 1\) for \(r > m\)

\( \Rightarrow  \qquad {T_r} < {T_{r + 1}}\)   for  \(1 \le r \le m\)    and \({T_r} > {T_{r + 1}}\) for \(r > m\)

\( \Rightarrow  \qquad {T_m} < {T_{m + 1}},\,\,{T_{m + 1}} > {T_{m + 2}}\)

\( \Rightarrow  \qquad {T_{m + 1}}\)   is the greatest term.

Example –5

How will you expand the multinomial expression \({\left( {{x_1} + {x_2} + ... + {x_m}} \right)^n}\) ?

Solution:    We will approach this problem using combinatorics. Note that a general term of the expansion would be of the form (without the coefficient)

\[x_1^{{n_1}}x_2^{{n_2}}x_3^{n{ \qquad_3}}....x_m^{{n_m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\]

where the various powers must always sum to n (why?).

i.e.,

\[{n_1} + {n_2} + {n_3} + ... + {n_m} = n\]

Now, to evaluate the coefficient of the term in (1), we consider the multinomial expression in expanded form:

\[\underbrace {\left( {{x_1} + {x_2} + ... + {x_m}} \right)\left( {{x_1} + {x_2} + ... + {x_m}} \right)..............\left( {{x_1} + {x_2} + ...{x_m}} \right)}_{n\,\,{\rm{times}}}\]

To generate the term in (1), we must get x 1  from n 1  terms, x 2  from n 2  terms and so on. Let us find the number of ways in which this can be done.

First select those n 1  multinomials that will contribute x 1  : this can be done in \(^n{C_{{n_1}}}\)  ways. Now, from the remaining \(\left( {n - {n_1}} \right)\)  multinomials, select those n 2  multinomials that will contribute x 2 : this can be done in \(^{\left( {n - {n_1}} \right)}{C_{{n_2}}}\) ways. Continuing this process, we see that the number of ways to get \({x_1}\,{\rm{from}}\,\,{n_1},{x_2}\) from n 2  ... and so on, that is, the number of times the term in (1) will be generated in the expansion is

\[\begin{align}{}\,\,\,\,\,&{\,^n}{C_{{n_1}}} \times {\,^{\left( {n - {n_1}} \right)}}{C_{{n_2}}} \times {\,^{\left( {n - {n_1} - {n_2}} \right)}}{C_{{n_3}}} \times ...\\ &= \frac{{n!}}{{{n_1}!\left( {n - {n_1}} \right)!}}\,\, \times \,\,\frac{{\left( {n - {n_1}} \right)!}}{{{n_2}!\left( {n - {n_1} - {n_2}} \right)!}}\,\, \times \,\,\frac{{\left( {n - {n_1} - {n_2}} \right)!}}{{{n_3}!\left( {n - {n_1} - {n_2} - {n_3}} \right)!}}\,\, \times \,\,...\\ &= \,\,\frac{{n!}}{{{n_1}!\,\,{n_2}!...{n_m}!}}\,\end{align}\]

This is what is known as the general multinomial coefficient . The multinomial expansion can now be written compactly as

\[{\left( {{x_1} + {x_2} + ... + {x_m}} \right)^n} = \sum {\frac{{n!}}{{{n_1}!{n_2}!...{n_m}!}}} \,\,x_1^{{n_1}}x_2^{{n_2}}...\,x_m^{{n_m}}\]

where the summation is carried out over all possible combinations of the n i 's such that \(\sum {{n_i} = n} \) . For example, in \({\left( {{x_1} + {x_2} + {x_3}} \right)^4}\,\) , let us consider some terms in the expansion:

Example –6

Find the coefficient of x 4  in the expansion of \({\left( {1 + x - \frac{2}{{{x^2}}}} \right)^{10}}\) .

Solution:    From the previous example, the general term in the expansion will be

\[\begin{align}{}\,\,\,\,\,\frac{{10!}}{{{n_1}!\;{n_2}!\;{n_3}!}}\;{(1)^{{n_1}}}{(x)^{{n_{^2}}}}{\left( {\frac{{ - 2}}{{{x^2}}}} \right)^{{n_{^3}}}}\\ = \frac{{10}}{{{n_1}!\;{n_2}!\;{n_3}!}}\;{x^{{n_2} - 2{n_3}}}{( - 2)^{{n_3}}}\end{align}\]

where \({n_1} + {n_2} + {n_3}\) must be 10.

Now, x 4  is generated whenever \({n_2} - 2{n_3} = 4.\) The possible values of the triplet \(({n_1},\;{n_2},\;{n_3})\) can now simply be listed out:

\[({n_1},\;{n_2},\;{n_3}) \equiv (6,\;4,\;0),\;(3,\;6,\;1),\;(0,\;8,\;2)\]

Thus, the (total) coefficient of x 4  is

\[\frac{{10!}}{{6!\;4!\;0!}}{( - 2)^0} + \frac{{10!}}{{3!\;6!\;1!}}{( - 2)^1} + \frac{{10!}}{{0!\;8!\;2!}}{( - 2)^2}\]

= – 1290 (verify)

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