ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 4- Worked Out Examples 2

Saturday, 9 August 2014

CHAPTER 4- Worked Out Examples 2

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: 4-(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:

	$q = \dfrac{{^n{C_{r + 1}}}}{{^n{C_r}}} = \dfrac{{\dfrac{{n!}}{{\left( {r + 1} \right)!\left( {n - r - 1} \right)!}}}}{{\dfrac{{n!}}{{r!\left( {n - r} \right)!}}}}$
	$= \dfrac{{n - r}}{{r + 1}}$

Thus

	$q > 1 \,\,\,\, \Rightarrow \,\,\,\, \dfrac{{n - r}}{{r + 1}} > 1$
	$\Rightarrow \,\,\,\, n - r > r + 1$
	$\Rightarrow \,\,\,\, r < \dfrac{{n - 1}}{2}$

Similarly,

	$q < 1 \Rightarrow \,\,\,\, \dfrac{{n - r}}{{r + 1}} < 1$
	$\Rightarrow \,\,\,\,\, n - r < r + 1$
	$\Rightarrow \,\,\,\, r > \dfrac{{n - 1}}{2}$

Thus,

$\left. \begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,\,\,{ ^n}{C_{r + 1}} \,\,>\,\, {\,^n}{C_r} \,\,\,\,\, {\rm{whenever}} \,\,\,\,\, r < \dfrac{{n - 1}}{2}\\ {\rm{and}} \,\,\, { ^n}{C_{r + 1}} \,\,< \,\,{\,^n}{C_r}\,\,\,\,\, {\rm{whenever}} \,\,\,\,\,r > \dfrac{{n - 1}}{2}\, \end{array} \right\}$

$\ldots(1)$

If $n$ is odd, we have

	$^n{C_{\dfrac{{n - 1}}{2}}} > {\,^n}{C_{\dfrac{{n - 3}}{2}}}$
	and $^n{C_{\dfrac{{n + 3}}{2}}} < {\,^n}{C_{\dfrac{{n + 1}}{2}}}$

Also, since

$^n{C_{\dfrac{{n - 1}}{2}}} = {\,^n}{C_{\dfrac{{n + 1}}{2}}}$

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

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

If $n$ is even, $(1)$ gives

	$^n{C_{\dfrac{n}{2}}} > {\,^n}{C_{\dfrac{n}{2} - 1}}$
	and $^n{C_{\dfrac{n}{2} + 1}} < \,{\,^n}{C_{\dfrac{n}{2}}}$

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

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

Solution: 4-(b)

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

	$q = \dfrac{{{T_{r + 1}}}}{{{T_r}}} = \dfrac{{^n{C_r}{x^{n - r}}{y^r}}}{{^n{C_{r - 1}}{x^{n - r + 1}}{y^{r - 1}}}}$
	$= \dfrac{{\left( {n - r + 1} \right)}}{r}.\dfrac{y}{x}$

Observe that

	$q > 1 \,\,\,\, \Rightarrow \,\,\,\, \dfrac{{\left( {n - r + 1} \right)}}{r} \cdot \dfrac{y}{x} > 1$
	$\Rightarrow \,\,\,\, \left\{ {\dfrac{{\left( {n + 1} \right)y}}{{x + y}} - r} \right\} \cdot \left( {\dfrac{{x + y}}{{r\,x}}} \right) > 0$

If $\dfrac{{\left( {n + 1} \right)y}}{{x + y}}$ 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

	$q > 1$ for $1 \le r < m$ and $q < 1$ for $r > m$
	$\Rightarrow \,\,\,\, {T_r} < {T_{r + 1}}$ for $1 \le r < m$ and ${T_r} > {T_{r + 1}}$ for $m + 1 \le r \le n$
	$\Rightarrow \,\,\,\, {T_{m - 1}} < {T_m}\,\,,\,\,{T_{m + 1}} > {T_{m + 2}},\,\,{T_m} = {T_{m + 1}}$
	$\Rightarrow \,\,\,\, {T_m}\,\,{\rm{and}}\,\,{T_{m + 1}}$ are the two greatest terms

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

We now have

	$q > 1$ for $1 \le r \le m$ and $q < 1$ for $r > m$
	$\Rightarrow \,\,\, {T_r} < {T_{r + 1}}$ for $1 \le r \le m$ and ${T_r} > {T_{r + 1}}$ for $r > m$
	$\Rightarrow \,\,\,\, {T_m} < {T_{m + 1}},\,\,{T_{m + 1}} > {T_{m + 2}}$
	$\Rightarrow \,\,\,\, {T_{m + 1}}$ is the greatest term

Saturday, 9 August 2014

CHAPTER 4- Worked Out Examples 2

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