ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 10

Saturday, 9 August 2014

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

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 $1$

Saturday, 9 August 2014