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Saturday, 9 August 2014

CHAPTER 10 - Worked Out Examples

     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} +\ldots \infty )^{ - n}}
Solution: 11-(a)

We have
{V_r} = \dfrac{{( - 3)\;(( - 3) - 1)(( - 3) - 2)) \ldots  (( - 3) - r + 1)}}{{r!}}
 = \dfrac{{{{( - 1)}^r}\;(r + 2)!}}{{2(r!)}}
 = \dfrac{{{{( - 1)}^r}\;(r + 1)\;(r + 2)}}{2}
Thus,
{V_0} = 1,\;\;{V_1} =  - 3,\;{V_2} = 6,\;{V_3} =  - 10\ldots
so that
{(1 - x)^{ - 3}} = 1 + 3x + 6{x^2} + 10{x^3} + \ldots \infty
Solution: 11-(b)

We use the result of part – (a)  in this:
{(1 + 3x + 6{x^2} + 10{x^3}\ldots \infty )^{ - n}} = {\left( {{{(1 - x)}^{ - 3}}} \right)^{ - n}}
 = {(1 - x)^{3n}}
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 = \dfrac{1}{8}.
Solution: 12

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,
{T_{r + 1}} = {V_r}\;{x^r}
and {T_r} = {V_{r - 1}}\;{x^{r - 1}}
so that \dfrac{{{T_{r + 1}}}}{{{T_r}}} = \dfrac{{{V_r}}}{{{V_{r - 1}}}} \cdot x
 = \dfrac{{n - r + 1}}{r} \cdot x
Now, let us find the conditions for which this ratio exceeds 1. We have
\left| {\;{T_{r + 1}}\;} \right|\; \ge \;\left| {\;{T_r}\;} \right|
 \Rightarrow \left| {\;\dfrac{{n + 1}}{r} - 1} \right|\; \ge \;\dfrac{1}{{|x|}}\ldots(1)
For this particular problem, (1)  becomes
\left| {\;\dfrac{{\dfrac{{ - 2}}{7} + 1}}{r} - 1} \right|\; \ge \;\dfrac{1}{{\left| {\dfrac{{ - 5}}{8}} \right|}}
 \Rightarrow\,\,\,\, \left| {\;\dfrac{5}{{7r}} - 1\;} \right|\; \ge \;\dfrac{8}{5}
 \Rightarrow \,\,\,\,\dfrac{5}{{7r}}\; \ge \;\dfrac{{13}}{5}
 \Rightarrow\,\,\,\, r \le \;\dfrac{{25}}{{91}}
\Rightarrow\,\,\,\, r=0
Thus, {T_{r + 1}} = {T_1} is the greatest term, with magnitude 1
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