Applications Of Calculus To Binomial Theorem
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
\[S = {\;^n}{C_1} + 2{\;^n}{C_2} + 3{\;^n}{C_3} + ...... + n{\;^n}{C_n}\]
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,
\[\begin{array}{l}S = {C_1} + 2{C_2} + ..... + {\;^n}{C_n}\\\,\,\,\, = \sum \;r\,{C_r}\end{array}\]
This series can be generated using a manipulation involving differentiation, as follows:
Consider the binomial expansion
\[{(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ...... + {C_n}{x^n}\]
If we differentiate both sides with respect to x, look at what we’ll obtain:
\[n{(1 + x)^{n - 1}} = {C_1} + 2{C_2}x + 3{C_3}{x^2} + ..... + n{C_n}{x^{n - 1}}\]
Now, all that remains is to substitute x = 1, upon which we obtain:
\[n \cdot {2^{n - 1}} = {C_1} + 2{C_2} + 3{C_3} + ..... + \;n\,{C_n}\]
This is what we were looking for. Thus,
Had we substituted \(x = - 1\) , we would’ve obtained
\[0 = {C_1} - 2{C_2} + 3{C_3}....... + {( - 1)^{n - 1}}\;n\,{C_n}\]
Thus, we have evaluated another interesting sum.
Suppose that we now wish to evaluate S 1 given by
\[{S_1} = {C_0} + \frac{{{C_1}}}{2} + \frac{{{C_2}}}{3} + .... + \frac{{{C_n}}}{{n + 1}}\]
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.
\[{(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ..... + {C_n}{x^n}\]
If we integrate this with respect to x, between some limits say a to b, we obtain
\[\left. {\frac{{{{(1 + x)}^{n + 1}}}}{{n + 1}}} \right|_a^b = \left. {{C_0}x} \right|_a^b + \left. {{C_1}\frac{{{x^2}}}{2}} \right|_a^b + \left. {{C_2}\frac{{{x^3}}}{3}} \right|_a^b + .... + {C_n}\left. {\frac{{{x^{n + 1}}}}{{n + 1}}} \right|_a^b\]
To generate the sum a little thought will show that we need to use a = 0, b = 1, so that we obtain
\[\frac{{{2^{n + 1}} - 1}}{{n + 1}} = {C_0} + \frac{{{C_1}}}{2} + \frac{{{C_2}}}{3} + ..... + \frac{{{C_n}}}{{n + 1}}\]
Thus, S 1 equals \(\begin{align}\frac{{{2^{n + 1}} - 1}}{{n + 1}}\end{align}\)
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.
Example – 7
Find the sum S given by
\[S = {1^2} \cdot {C_1} + {2^2} \cdot {C_2} + {3^2} \cdot {C_3} + .... + {n^2} \cdot {C_n}\]
Solution: 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:
\[{(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ...... + {C_n}{x^n}\]
Differentiating both sides with respect to x, we have
\[n{(1 + x)^{n - 1}} = {C_1} + 2{C_2}x + 3{C_3}{x^2} + ...... + n{C_n}{x^{n - 1}}\]
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:
\[nx{(1 + x)^{n - 1}} = {C_1}x + 2{C_2}{x^2} + 3{C_3}{x^3} + .... + n{C_n}{x^n}\]
It should be evident now that the next step is differentiation:
\[n(n - 1)x{(1 + x)^{n - 2}} + n{(1 + x)^{n - 1}} = {C_1} + {2^2} \cdot {C_2}x + {3^2} \cdot {C_3}{x^2} + .... + {n^2} \cdot {C_n}{x^{n - 1}}\]
Now we simply substitute x = 1 to obtain
\[n(n - 1) \cdot {2^{n - 2}} + n \cdot {2^{n - 1}} = {C_1} + {2^2} \cdot {C_2} + {3^2} \cdot {C_3} + ..... + {n^2} \cdot {C_n}\]
The required sum S is thus
\[\begin{array}{l}S\;\; = \;n(n - 1) \cdot {2^{n - 2}} + n \cdot {2^{n - 1}}\\\\\,\,\,\,\, = n \cdot {2^{n - 2}}\left\{ {(n - 1) + 2} \right\}\\\\\,\,\,\,\, = n(n + 1) \cdot {2^{n - 2}}\end{array}\]
Example – 8
Evaluate the following sums:
(a) \(\begin{align}{S_1} = \frac{{{C_0}}}{1} + \frac{{{C_2}}}{3} + \frac{{{C_4}}}{5} + .......\end{align}\) (b) \(\begin{align}{S_2} = \frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .......\end{align}\)
Solution: 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
\[S = {C_0} + \frac{{{C_1}}}{2} + \frac{{{C_2}}}{3} + ...... + \frac{{{C_n}}}{{n + 1}} = \frac{{{2^{n + 1}} - 1}}{{n + 1}}\]
Note that S is the sum of S 1 and S 2 , i.e.,
\[{S_1} + {S_2} = \frac{{{2^{n + 1}} - 1}}{{n + 1}}\]
Thus, if we determine S 1 , S 2 is automatically determined, and vice-versa. Let us try to determine S 1 first.
(a) Consider again the general expansion
\[{(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}\]
Integrating with respect to x, we have (we have not yet decided the limits)
\[\left. {\frac{{{{(1 + x)}^{n + 1}}}}{{n + 1}}} \right|_a^b = \left. {{C_0}x} \right|_a^b + \left. {{C_1}\frac{{{x^2}}}{2}} \right|_a^b + \left. {{C_2}\frac{{{x^3}}}{3}} \right|_a^b + ..... + \left. {{C_n}\frac{{{x^{n + 1}}}}{{n + 1}}} \right|_a^b\]
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
\[\frac{{{2^{n + 1}}}}{{n + 1}} = 2\left( {{C_0} + \frac{{{C_2}}}{3} + \frac{{{C_4}}}{5} + ....} \right)\]
which implies that
\[{S_1} = \frac{{{2^n}}}{{n + 1}}\]
(b) S 2 is now simply given by
\[\begin{align}{}{S_2}\;\; &= S - {S_1}\\\,\,\,\,\,\,\, &= \frac{{{2^{n + 1}} - 1}}{{n + 1}} - \frac{{{2^n}}}{{n + 1}}\\\,\,\,\,\,\,\, &= \frac{{{2^n} - 1}}{{n + 1}}\\\end{align}\]