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Saturday, 9 August 2014

CHAPTER 7 - Worked Out Examples

 Example: 7     

Find the sum S given by
S = {1^2} \cdot {C_1} + {2^2} \cdot {C_2} + {3^2} \cdot {C_3} + \ldots + {n^2} \cdot {C_n}
Solution: 7

We have to plan an approach wherein we are able to generate {r^2} with {C_r}. We can generate one r with every {C_r}, as we did earlier, and which is now repeated here:
{(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + \ldots  + {C_n}{x^n}
Differentiating both sides with respect to x, we have
n{(1 + x)^{n - 1}} = {C_1} + 2{C_2}x + 3{C_3}{x^2} + \ldots + n{C_n}{x^{n - 1}}
Now we have reached the stage where we have an r with every {C_r}. We need to think how to get the other r. If we differentiate once again, we’ll have r(r - 1) with every {C_r} instead of {r^2} (understand this point carefully). To ‘make-up’ for the power that falls one short of the required value, we simply multiply by x on both sides of the relation above to obtain:
nx{(1 + x)^{n - 1}} = {C_1}x + 2{C_2}{x^2} + 3{C_3}{x^3} + \ldots  + n{C_n}{x^n}
It should be evident now that the next step is differentiation:
n(n - 1)x{(1 + x)^{n - 2}} + n{(1 + x)^{n - 1}} = {C_1} + {2^2} \cdot {C_2}x + {3^2} \cdot {C_3}{x^2} +\ldots + {n^2} \cdot {C_n}{x^{n - 1}}
Now we simply substitute x = 1  to obtain
n(n - 1) \cdot {2^{n - 2}} + n \cdot {2^{n - 1}} = {C_1} + {2^2} \cdot {C_2} + {3^2} \cdot {C_3} +\ldots + {n^2} \cdot {C_n}
The required sum S is thus
S\;\; = \;n(n - 1) \cdot {2^{n - 2}} + n \cdot {2^{n - 1}}
 = n \cdot {2^{n - 2}}\left\{ {(n - 1) + 2} \right\}
 = n(n + 1) \cdot {2^{n - 2}}
     Example: 8     

Evaluate the following sums:
(a) {S_1} = \dfrac{{{C_0}}}{1} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} +\ldots
(b) {S_2} = \dfrac{{{C_1}}}{2} + \dfrac{{{C_3}}}{4} + \dfrac{{{C_5}}}{6} +\ldots
Solution: 8-(a)

The first sum contains only the even-numbered binomial coefficients, while the second contains only odd-numbered ones. Recall that we have already evaluated the sum S given by
S = {C_0} + \dfrac{{{C_1}}}{2} + \dfrac{{{C_2}}}{3} + \ldots + \dfrac{{{C_n}}}{{n + 1}} = \dfrac{{{2^{n + 1}} - 1}}{{n + 1}}
Note that S is the sum of {S_1} and {S_2}, i.e.,
{S_1} + {S_2} = \dfrac{{{2^{n + 1}} - 1}}{{n + 1}}
Thus, if we determine {S_1} , {S_2} is automatically determined, and vice-versa. Let us try to determine {S_1} first.
Consider again the general expansion
{(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} +\ldots + {C_n}{x^n}
Integrating with respect to x, we have (we have not yet decided the limits)
\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 + \left. {{C_n}\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right|_a^b
Since we are trying to determine {S_1} which contains only the even-numbered terms, we have to choose the limits of integration such that the odd-numbered terms vanish. This is easily achievable by setting a =  - 1 and b = 1  (understand this carefully). Thus, we have
\dfrac{{{2^{n + 1}}}}{{n + 1}} = 2\left( {{C_0} + \dfrac{{{C_2}}}{3} + \dfrac{{{C_4}}}{5} + \ldots } \right)
which implies that
{S_1} = \dfrac{{{2^n}}}{{n + 1}}
Solution: 8-(b)

{S_2} is now simply given by
{S_2}\;\; = S - {S_1}
 = \dfrac{{{2^{n + 1}} - 1}}{{n + 1}} - \dfrac{{{2^n}}}{{n + 1}}
 = \dfrac{{{2^n} - 1}}{{n + 1}}
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