Binomial Theorem
10.0 Use of calculus
10.0 Use of calculus
The questions in which the relation between binomial coefficients is to be proved, can be solved either using the properties of binomial coefficient or using calculus i.e., differentiation and integration which makes it easier to find the relation among the binomial coefficients.
1. Differentiation: This method is applied only when the numericals occur as the product of the binomial coefficients.
Differentiating ${(1 + x)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}$ on both sides, we get $$n{(1 + x)^{n - 1}} = {}^n{C_1} + 2{}^n{C_2}x + 3{}^n{C_3}{x^2} + ... + n{}^n{C_n}{x^{n - 1}}\quad ...(1)$$
Put $x=1$ in $(1)$, we get $$n{(2)^{n - 1}} = {}^n{C_1} + 2{}^n{C_2} + 3{}^n{C_3} + ... + n{}^n{C_n}$$
Put $x=-1$ in $(1)$, we get $$0 = {}^n{C_1} - 2{}^n{C_2} + 3{}^n{C_3} - ... + {( - 1)^{n - 1}}n{}^n{C_n}$$
On differentiating $(1)$ again and again we will have different results.
2. Integration: This method is applied only when the numericals occur as the denominator of the binomial coefficient.
Integrating ${(1 + x)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}$ on both sides, we get
$$\frac{{{{(1 + x)}^{n + 1}}}}{{n + 1}} + {\text{constant (}}c'){ = ^n}{C_0}x + \frac{{^n{C_1}{x^2}}}{2} + \frac{{^n{C_2}{x^3}}}{3} + ... + \frac{{^n{C_n}{x^{n + 1}}}}{{n + 1}}...(2)$$
For $x=0$, we get $c' = - \frac{1}{{n + 1}}$. Put the value of $c'$ in equation $(1)$ $$\frac{{{{(1 + x)}^{n + 1}} - 1}}{{n + 1}}{ = ^n}{C_0}x + \frac{{^n{C_1}{x^2}}}{2} + \frac{{^n{C_2}{x^3}}}{3} + ... + \frac{{^n{C_n}{x^{n + 1}}}}{{n + 1}}...(3)$$
Put $x=1$ in equation $(3)$, we get $$\frac{{{{(2)}^{n + 1}} - 1}}{{n + 1}}{ = ^n}{C_0} + \frac{{^n{C_1}}}{2} + \frac{{^n{C_2}}}{3} + ... + \frac{{^n{C_n}}}{{n + 1}}$$
Put $x=-1$ in equation $(3)$, we get $$\frac{1}{{n + 1}} = {}^n{C_0} - \frac{{{}^n{C_1}}}{2} + \frac{{{}^n{C_2}}}{3} - ...$$
On integrating $(3)$ again and again we will have different results.
Question 10. If ${(1 + x)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}} {x^r}$ then prove that $${}^n{C_0} + 2{}^n{C_1} + 3{}^n{C_2} + ... + (n + 1){}^n{C_n} = {2^{n - 1}}(n + 2)$$
Solution: We have $${(1 + x)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}\quad ...(1)$$
Multiply $(1)$ with $x$, we get
$$x{(1 + x)^n} = {}^n{C_0}x + {}^n{C_1}{x^2} + {}^n{C_2}{x^3} + ... + {}^n{C_n}{x^{n + 1}}\quad ...(2)$$
Differentiating $(2)$ with respect to $x$, we get $${(1 + x)^n} + n{(1 + x)^{n - 1}}x = {}^n{C_0} + 2{}^n{C_1}x + 3{}^n{C_2}{x^2} + ... + (n + 1){}^n{C_n}{x^n}$$
Put $x=1$, we get $$\begin{equation} \begin{aligned} {(2)^n} + n{(2)^{n - 1}} = {}^n{C_0} + 2{}^n{C_1} + 3{}^n{C_2} + ... + (n + 1){}^n{C_n} \\ {2^{n - 1}}(n + 2) = {}^n{C_0} + 2{}^n{C_1} + 3{}^n{C_2} + ... + (n + 1){}^n{C_n} \\\end{aligned} \end{equation} $$
