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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
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