Binomial Theorem For Any Index

BINOMIAL THEOREM FOR ANY INDEX

(1+x)ⁿ = 1+ nx + $\frac{n(n-1)}{2!}{x}^2$ + . . . + $\frac{n(n-1)\cdots (n-r+1)}{r!}{x}^r+\cdots termsupto\infty$

Observations:

$\bullet$Expansion is valid only when –1 <x <1

$\bullet$General term of the series (1+x)^-n = T_r+1 = (-1)^r $\frac{n(n+1)(n+2)\cdots (n+r-1)}{r!}$x^r

$\bullet$General term of the series (1-x)^-n = T_r+1 = $\frac{n(n+1)(n+2)\cdots (n+r-1)}{r!}$x^r

$\bullet$If first term is not 1, then make first term unity in the following way:

(a+ x)ⁿ = aⁿ(1+x/a)ⁿ if $\left|\frac{x}{a}\right|....<1$

IMPORTANT EXPANSIONS

$\bullet$(1+ x)^-1 = 1- x +x² –x³ + . . . + (-1)^rx^r+. . .

$\bullet$(1 - x)^-1 = 1+ x +x² +x³ + . . .+ x^r + . . .

$\bullet$(1+ x)^-2 = 1- 2x +3x² –4x³+ . . .+ (-1)^r(r+1)x^r+. . .

$\bullet$(1 - x)^-2 = 1+ 2x +3x² +4x³ + . . .+ (r+1)x^r+. . .

$\bullet$(1+x)^-3 = 1- 3x +6x² –10x³ +. . .+ (-1)^r $\frac{\left(r+1\right)\left(r+2\right)}{r!}{x}^r....1$

$\bullet$(1-x)^-3 = 1+ 3x +6x² +10x³ + . . .+ $\frac{\left(r+1\right)\left(r+2\right)}{r!}{x}^r+..$ .

$\bullet$In general coefficient of x^r in (1 – x)^{– n} is ^{n + r –1}C_r .

$\bullet$(1 – x)^–p/q = 1 + $\frac{p}{1!}\left(\frac{x}{q}\right)+\frac{p\left(p+q\right)}{2!}{\left(\frac{x}{q}\right)}^2-$ + …….

$\bullet$(1 + x)^–p/q = 1 – $\frac{p}{1!}\left(\frac{x}{q}\right)+\frac{p\left(p+q\right)}{2!}{\left(\frac{x}{q}\right)}^2-$ – …….

$\bullet$(1 + x)^p/q = 1 + $\frac{p}{1!}\left(\frac{x}{q}\right)+\frac{p\left(p-q\right)}{2!}{\left(\frac{x}{q}\right)}^2$ + …….

$\bullet$(1 – x)^p/q = 1 – $\frac{p}{1!}\left(\frac{x}{q}\right)+\frac{p\left(p-q\right)}{2!}{\left(\frac{x}{q}\right)}^2$ – …….

Illustration 1: If –1 < x < 1, show that (1 –x)^-2 = 1 + 2x + 3x² + 4x³ + …..to $\infty$.

Solution: We know that if n is a negative integer or fraction

(1+x)ⁿ= 1 + $\frac{n}{1!}x+\frac{n\left(n-1\right)}{2!}{x}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^3+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{x}^4......to\infty$

Provided –1 < x < 1

Putting n = -2 and –x in place of x, we get

(1+x)² = 1 + $\frac{\left(-2\right)}{1!}\left(-x\right)+\frac{\left(-2\right)\left(-2-1\right)}{2!}{\left(-x\right)}^2+\frac{\left(-2\right)\left(-2-1\right)\left(-2-2\right)}{3!}{\left(-x\right)}^3......to\infty$

= 1 + 2x + 3x² + 4x³ + … to ¥.

MULTINOMIAL EXPANSION

In the expansion of (x₁+x₂ + . . . + x_n)^m where m, n $\in$ N and x₁, x₂ , . . ., x_n are independent variables, we have

$\bullet$Total number of term in the expansion = ^m+n-1C_n-1

$\bullet$Coefficient of ${x_1}^{r_1}{x_2}^{r_2}{x_3}^{r_3}\cdots {x_n}^{r_n}$(where r₁ + r₂ +…+ r_n = m, r_i $\in$N $\cup$ {0} is $\frac{m!}{r_1!r_2!\cdots r_n!}$.

$\bullet$ Sum of all the coefficient is obtained by putting all the variables x_i equal to 1 and it is equal to n^m.

Illustration 2: If x₁ + x₂ + x₃ + x₄ + x₅ = 20 and x₁ + x₂ = 5 , (x₁ ,x₂ , x₃ ,x₄ , x₅ ³ 0) then find the number of non negative integral solutions of above equation.

Solution: x₁ + x₂ + x₃ + x₄ + x₅ = 20 , x₁ + x₂ = 5 … (1)

$\Rightarrow$ x₃ + x₄ + x₅ = 15 … (2)

Number of solutions

$\Rightarrow$Coefficient of x⁵ in (1) x coefficient of x¹⁵ in (2)

$\Rightarrow$Coefficient of x⁵ ${\left(\frac{1-x^6}{1-x}\right)}^2$x Coefficient of x¹⁵ ${\left(\frac{1-x^{16}}{1-x}\right)}^3$

$\Rightarrow$Coefficient of x⁵ in (1 – x)^-2 x Coefficient of x¹⁵ ${(1-x)}^{-3}$

^{2 + 5 – 1}C₁ x ^3+15-1C_3-1 = ⁶C₁ x ¹⁷C₂ = $\frac{6\times 17\times 16}{2}=48\times 17=816$