Properties Of Binomial Coefficient

PROPERTIES OF BINOMIAL COEFFICIENT

For the sake of convenience the coefficients ⁿC₀ , ⁿC₁ , . . ., ⁿC_r , . . . ,ⁿC_n are usually denoted by C₀, C₁ , . . . , C_r , . . . ,C_n respectively

$\bullet$C₀ + C₁ + C₂ +. . . . . + C_n = 2ⁿ

$\bullet$C₀ - C₁ + C₂ -. . . . . + (–1)ⁿ C_n = 0

$\bullet$C₀ + C₂ + C₄ +. . . . . = C₁ + C₃ + C₅ +. . . . . = 2^n-1

$\bullet$ ${}^n{C}_{r_1}{=}^n{C}_{r_2}$ $\Rightarrow$ r₁ = r₂ or r₁ + r₂ = n

^$\bullet$nC_r + ⁿC_r-1 = ⁿ⁺¹C_r

$\bullet$r ⁿC_r =n ^n-1C_r-1

$\bullet$$\frac{{}^n{C}_r}{r+1}=\frac{{}^{n+1}{C}_{r+1}}{n+1}$.Illustration 1:Find the value of $\sum _{r=0}^n\left(\frac{r+2}{r+1}\right){}^n{C}_r$Solution:The given value is $\sum _{r=0}^n\left(\frac{r+2}{r+1}\right){}^n{C}_r=\sum _{r=0}^n\left(1+\frac{1}{r+1}\right){}^n{C}_r$ $=\sum _{r=0}^n{}^n{C}_r+\frac{1}{n+1}\sum _{r=0}^n{}^{n+1}{C}_{r+1}$$=2^n+\frac{1}{n+1}(2^{n+1}-1)$ $=\frac{2^n(n+3)-1}{n+1}$

PROBLEMS RELATED TO SERIES OF BINOMIAL COEFFICIENTS

$\bullet$Problems involving binomial coefficients with alternate sign:

Illustration 2: Evaluate C₀ - C₁ + C₂ - C₃ +...+ (-1)ⁿC_n.

Solution: Here alternately +ve and - ve sign occur

This can be obtained by putting (-1) instead of 1 in place of x in

(1 + x)ⁿ = C₀ + C₁x +...+ ⁿC_nxⁿ, we get C₀ - C₁ +...+ (-1)ⁿC_n = 0

Now to obtain the sum C₀ + C₂ + C₄ + ...

we add (1 + 1)ⁿ and (1 - 1)ⁿ.

Similarly, the cube roots of unity may be used to evaluate

C₀ + C₃ + C₆ + ... OR C₁ + C₄ +... OR C₂ + C₅ +...

put x = 1, x = w, x = w² in

(1 + x)ⁿ = C₀ + C₁x +...+ C_nxⁿ and add to get C₀ + C₃ + C₆ +...

the other two may be obtained by suitably multiplying (1 + w)ⁿ and

(1 + w²)ⁿ by w and w² respectively.

$\bullet$Problems Related to series of Binomial coefficients in which each term is a product of an integer and a binomial coefficient i.e. in the form k ⁿC_r

Illustration 3: Prove that $\frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+....=\frac{2^n}{n+1}$

Solution: Consider the expansion

(1 + x)ⁿ = C₀ + C₁x + C₂x² + C₃x³ + C₄x⁴ + … + C_nxⁿ …(i)

Integrating both sides of (i) within limits –1 to 1, we get

${\int }_{-1}^1{(1+x)}^{n}dx={\int }_{-1}^1(C_0+C_{1}x+C_2{x}^2+C_3{x}^3+C_4{X}^4+...C_n{x}^n)dx$

= ${\int }_{-1}^1(C_0+C_2{x}^2+C_4{x}^4+...)dx{\int }_{-1}^1(C_{1}x+C_3{x}^3+....)dx$

= $2{\int }_{-1}^1(C_0+C_2{x}^2+C_4{x}^4+...)dx+0$ (By Prop. Of definite integral)(since second integral contains odd function)

$\frac{{(1+x)}^{n+1}}{n+1}={{\left]{}_{+1}^1=2\left(C_{0}x+\frac{C_2{x}^3}{3}+\frac{C_4{X}^5}{5}+...\right)\right]}^1}_0$

$\frac{2^{n+1}}{n+1}=2\left(C_0\frac{C_2}{3}+\frac{C_4}{5}+...\right)$

Hence $C_0\frac{C_2}{3}+\frac{C_4}{5}+...=\frac{2^n}{n+1}$

Alternative Method.

L.H.S. = C0 = $\frac{C_2}{3}+\frac{C_4}{5}+...$

= 1 + $\frac{n(n-1)}{\mathrm{1.2.3}}+\frac{n(n-1)(n-2)(n-2)}{\mathrm{1.2.3.4.5}}+....$

= $\frac{1}{(n+1)}\left\{\frac{(n+1)}{1}+\frac{(n+1)n(n-1)}{\mathrm{1.2.3}}+\frac{(n+1)n(n-1)(n-2)(n-3)}{\mathrm{1.2.3.4.5}}+...\right\}$

= $\frac{1}{(n+1)}$ {ⁿ⁺¹C₁+ⁿ⁺¹C₃+ⁿ⁺¹C₅+…}

= $\frac{1}{(n+1)}$ {sum of even binomial coefficients of (1 + x)ⁿ⁺¹}

= $\frac{2^{n+1-1}}{(n+1)}$

= $\frac{2^n}{n+1}$ = R. H. S. coefficient divided by an integer i.e. in the form of $\frac{{}^n{C}_r}{k}$.

v Problem related to series of binomial coefficients in which each term is a product of two binomial coefficients.

Solution Process:

(1) If difference of the lower suffixes of binomial coefficients in each term is same.

i.e. C₁C₃ + C₂C₄ + C₃C₅ + …

Here 3 – 1 = 4 – 2 = 5 – 3 = … = 2

Case I: If each term of series is positive then

(1 + x)ⁿ = C₀ + C₁x + C₂x² + …. + C_nxⁿ …(i)

Interchanging 1 and x,

(x + 1)ⁿ = C₀xⁿ + C₁x^n–1 + C₂x^n–2 + …+ C_n …(ii)

Then multiplying (i) and (ii) and equate the coefficient of suitable power of x on both sides

Or

Replacing x by $\frac{1}{x}$ in (i), then

${\left(1+\frac{1}{x}\right)}^n=C_0+\frac{C_1}{x}+\frac{C_2}{x^2}+...+\frac{C_n}{x^n}$ ….(iii)

Then multiplying (i) and (ii) and equate the coefficient of suitable power of x on both sides.

Case II: If terms of the series alternately positive and negative then

(1–x)ⁿ = C₀ – C¹ x + C₂x² - … + (–1)ⁿC_nxⁿ …(i)

and (x +1)ⁿ = C₀xⁿ+C₁x^n–1 + C₂x^n–2 + … + Cn …(ii)

Then multiplying (i) and (ii) and equate the coefficient of suitable power of x on both sides.

Or

Replacing x by $\frac{1}{x}$ in (i), then ${\left(1-\frac{1}{x}\right)}^n=C_0+\frac{C_1}{x}+\frac{C_2}{x^2}+...+\frac{{(-1)}^n{C}_n}{x^n}$ …(iii)

Then multiplying (i) and (iii) and equate the coefficient of suitable power of x on both sides.

Illustration 4: If I is integral part of (2 + $\sqrt{3}$)ⁿ and f is fraction part of (2 + $\sqrt{3}$)ⁿ, then prove that (I + f) (1 –f) = 1. Also prove that I is an odd Integer.

Solution: (2 + $\sqrt{3}$)ⁿ = I + f where I is an integer and 0 £ f < 1

Here note that (2 - $\sqrt{3}$)ⁿ (2 +$\sqrt{3}$)ⁿ = (4 - 3)ⁿ = 1

Since (2 + $\sqrt{3}$)ⁿ (2 -$\sqrt{3}$)ⁿ = 1 it is thus required to prove that

(2 - $\sqrt{3}$)ⁿ = 1 - f

but, (2 - $\sqrt{3}$)ⁿ + (2 +$\sqrt{3}$)ⁿ = [2ⁿ - C₁.2^{n - 1}.$\sqrt{3}$ + C₂2^{n - 2.}.($\sqrt{3}$)² - ...]

+ [2ⁿ + C₁.2^{n - 1}.$\sqrt{3}$ + C₂2^{n -2.}.($\sqrt{3}$)² - ...]

= 2[2ⁿ + C₂.2^{n - 2}.3+C₄2^{n - 4}.3² + ...] = even integer

Now 0 < (2 - $\sqrt{3}$) < 1

0 < (2 - $\sqrt{3}$)ⁿ < 1

if (2 - $\sqrt{3}$)ⁿ = f ', then I + f + f ' = Even

Now O $⩽$f < 1 and 0 < f ' < 1 ……(1)

Also I + f + f ' = Even integer

f + f ' = integer ……(2)

(1) and (2) imply that f + f ' = 1 ( since 0 < f + f ' < 2)

$\Rightarrow$ I is odd and f ' = 1 - f $\Rightarrow$ (I + f) (1 - f) = 1.