ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 3

Saturday, 9 August 2014

CHAPTER 3 - Worked Out Examples

Example: 1

Find the middle term(s) in the expansion of ${\left( {x - \dfrac{1}{x}} \right)^9}$

Solution: 1

Since $n = 9$ , there will be $10$ terms in the expansion, which means that there will be $2$ middle terms in the expansion, the ${5^{th}}$ and the ${6^{th}}$ :

	${T_5} = {\,^9}{C_4}{\left( x \right)^{9 - 4}}\left( {\dfrac{{ - 1}}{x}} \right)_{}^4 = 126\,x$
	${T_6} = {\,^9}{C_5}{\left( x \right)^{9 - 5}}{\left( {\dfrac{{ - 1}}{x}} \right)_5} = \dfrac{{ - 126}}{x}$

Example: 2

Is there any term in the expansion of ${\left( {\dfrac{{2{x^2}}}{5} - \dfrac{1}{{\sqrt x }}} \right)^{10}}$ that will be independent of $x$ ?

Solution: 2

The general term in the expansion is

	${T_{r + 1}} = {\,^{10}}{C_r}{\left( {\dfrac{{2{x^2}}}{5}} \right)^{10 - r}}{\left( {\dfrac{{ - 1}}{{\sqrt x }}} \right)^r}$
	$= {\,^{10}}{C_r}{\left( {\dfrac{2}{5}} \right)^{10 - r}}{\left( { - 1} \right)^r}\,\,{x^{20 - 2r - \dfrac{r}{2}}}$
	$= {\,^{10}}{C_r}{\left( {\dfrac{2}{5}} \right)^{10 - r}}{\left( { - 1} \right)^r}\,\,{x^{20\, - \dfrac{{5r}}{2}}}$

Thus, for the term that is independent of $x$ , we have

	$20 - \dfrac{{5r}}{2} = 0$
	$\Rightarrow \,\,\,\, r = 8$

Thus, the term free of $x$ is the ${9^{th}}$ term given by

${T_9} = {\,^{10}}{C_8}{\left( {\dfrac{2}{5}} \right)^2}{\left( { - 1} \right)^8} = \dfrac{{36}}{5}$

Example: 3

Evaluate the sum $^n{C_0} + {\,^n}{C_1} + {\,^n}{C_2} + \ldots + {\,^n}{C_n}$ .

Solution: 3

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 ${\rm{2 \times 2 \times 2 }}\ldots n\,{\rm{ times = }}{{\rm{2}}^{\rm{n}}}$ . The sum of the binomial coefficients therefore equals ${2^n}$ .

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

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

If we put $x = 1$ , we simply obtain

${2^n} = {\,^n}{C_0} + {\,^n}{C_1} + {\,^n}{C_2} +\ldots {\,^n}{C_n}$

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

	$0 = {\,^n}{C_0} - {\,^n}{C_1} + {\,^n}{C_2} - {\,^n}{C_3} + \ldots + {\left( { - 1} \right)^n}{\,^n}{C_n}$
	$\Rightarrow \,\,\,\, { ^n}{C_0} + {\,^n}{C_2} + {\,^n}{C_4} + \ldots = {\,^n}{C_1} + {\,^n}{C_3} + {\,^n}{C_5} +\ldots$

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:

	$^n{C_0}{2^n} + {\,^n}{C_1}{2^{n - 1}} + {\,^n}{C_2}\,{2^{n - 2}} + \ldots {\,^n}{C_n} = {3^n}$
	$^n{C_0} - {\,^n}{C_1}x + {\,^n}{C_2}\,{x^2}\,\, - \ldots \, + {\left( { - 1} \right)^n}{\,^n}{C_n}{x^n} = {\left( {1 - x} \right)^n}$
	$^n{C_0} - \,\dfrac{{^n{C_1}}}{2} + \,\dfrac{{^n{C_2}}}{{{2^2}}}\, - \dfrac{{^n{C_3}}}{{{2^3}}} + \ldots \, + \dfrac{{{{\left( { - 1} \right)}^n}{\,^n}{C_n}}}{{{2^n}}} = \dfrac{1}{{{2^n}}}$

Saturday, 9 August 2014

CHAPTER 3 - Worked Out Examples

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