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Saturday, 9 August 2014

CHAPTER 6 - Binomial Involving Calculus

The techniques of calculus enable us to sum a lot of series involving binomial coefficients. This is the subject of this section.
Suppose that we have to evaluate the sum S  given by
S = {\;^n}{C_1} + 2{\;^n}{C_2} + 3{\;^n}{C_3} + \ldots  + n{\;^n}{C_n}
From now on, to avoid clutter, we’ll write ^n{C_r} as simply {C_r}, where the upper index n should be understood to be present. Thus,
S = {C_1} + 2{C_2} + \ldots + {\;^n}{C_n}
 = \sum \;r\,{C_r}
This series can be generated using a manipulation involving differentiation, as follows:
Consider the binomial expansion
{(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + \ldots  + {C_n}{x^n}
If we differentiate both sides with respect to x, look at what we’ll obtain:
n{(1 + x)^{n - 1}} = {C_1} + 2{C_2}x + 3{C_3}{x^2} +\ldots + n{C_n}{x^{n - 1}}
Now, all that remains is to substitute x = 1, upon which we obtain:
n \cdot {2^{n - 1}} = {C_1} + 2{C_2} + 3{C_3} + \ldots  + \;n\,{C_n}
This is what we were looking for. Thus, S = n \cdot {2^{n - 1}}
Had we substituted x =  - 1, we would’ve obtained
0 = {C_1} - 2{C_2} + 3{C_3}\ldots  + {( - 1)^{n - 1}}\;n\,{C_n}
Thus, we have evaluated another interesting sum.
Suppose that we now wish to evaluate {S_1} given by
{S_1} = {C_0} + \dfrac{{{C_1}}}{2} + \dfrac{{{C_2}}}{3} +\ldots + \dfrac{{{C_n}}}{{n + 1}}
The alert reader would immediately realize that integration needs to be applied here. How exactly to do so is now described. Consider again the expansion.
{(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} +\ldots + {C_n}{x^n}
If we integrate this with respect to x, between some limits say a to b, we obtain
\left. {\dfrac{{{{(1 + x)}^{n + 1}}}}{{n + 1}}} \right|_a^b = \left. {{C_0}x} \right|_a^b + \left. {{C_1}\dfrac{{{x^2}}}{2}} \right|_a^b + \left. {{C_2}\dfrac{{{x^3}}}{3}} \right|_a^b + \ldots  + {C_n}\left. {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right|_a^b
To generate the sum {S_1} a little thought will show that we need to use a = 0, b = 1, so that we obtain
\dfrac{{{2^{n + 1}} - 1}}{{n + 1}} = {C_0} + \dfrac{{{C_1}}}{2} + \dfrac{{{C_2}}}{3} + \ldots + \dfrac{{{C_n}}}{{n + 1}}
Thus, {S_1} equals \dfrac{{{2^{n + 1}} - 1}}{{n + 1}}
Try some other values for a and b and hence generate other series on your own. Be as varied as you can in choosing these limits.
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