Kenneth Rosen Edition 6th Exercise 6.4 Question 6 e (Page No. 440)

Question

nC2 for n=0,1,2,3...

Comments

Can u tell chapter name, division and number?

---

Advance Counting techniques , div 6 , Q-e

Answer 1

Series will be: 0, 0, 1, 3, 6, 10, 15, 21, ....
Encoded in power series$: 1x^2 + 3x^3 + 6x^4 + 10x^5 + 15x^6 + 21x^7 + ...$
let,

$\ \ \ f(x) \ \ \ = 1x^2 + 3x^3 + 6x^4 + 10x^5 + 15x^6 + 21x^7 + ...$
$x*f(x) =           1x^3 + 3x^4 + 6x^5 + 10x^6 + 15x^7 + 21x^8 + ...$

 

Subtract the equations, we get:

$(1 - x)*f(x) = x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + 6x^7 + ...$
$(1 - x)*f(x) = x^2 (1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + ... )$

$(1 - x)*f(x) = \frac{x^2}{(1 - x)^2}$  (Generating function for $1 + 2x + 3x^2 + ...$ is $1/(1-x)^2$)

$f(x) = \frac{x^2}{(1 - x)^3}$

So $\frac{x^2}{(1 - x)^3}$ is the closed form of the given series.

Comments

can you please provide link for such approach?

---

I learned it here and this method works almost on every sequence:
https://www.youtube.com/watch?v=Pp4PWCPzeQs

---

@neeraj33negi  can you please tell me way for 10 C n+1 , i do not want to remember  formula from rosen ...thanks

---

I have not used any such formula but you can easily prove the formula in 3 lines.
Just expand:
$C(n, k+1) = \frac{n!}{(k+1)!(n-k-1)!}$
=> $\frac{n!}{(k+1) (k)!(n-k-1)!}$

=> $\frac{n!(n-k)}{(k+1) (k)!(n-k-1)!(n-k)}$ (Multiply numerator and denominator by $(n-k)$)

=>$\frac{n!}{(k)!(n-k)!}$$\frac{n-k}{(k+1)}$

=>$C(n,k)\frac{(n-k)}{(k+1)}$
I hope this is what you were asking for.