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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}}}
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