What are the steps to solving this algebraic summation problem?

[Homewoodm01]

Homework Statement

Ok I have the answer to a question, all the working is given, however, I'm having trouble following it.

Homework Equations

http://img695.imageshack.us/img695/426/answer.jpg

The Attempt at a Solution

I am completely lost, could someone please explain the steps that have been taken, it would really help me.

 

[statdad]

First line to second: use the definition of \binom x y, then factor out terms that do not depend on x.

<br /> \binom x y= \frac{x!}{y! (x-y)!}<br />

and factor from the sum any term that does not depend on the index of summation, x

Second line to third: shift the origin of summation from y to 0 by replacing the index of summation by x = y. After this the sum becomes

<br /> \sum_{x=0}^\infty \lambda^{x+y} \frac{0.9^x}{x!} = \lambda^y \sum_{x=0}^\infty \frac{(0.9\, \lambda)^x}{x!}<br />

You should be able to fill in the final step yourself.

 

[Homewoodm01]

Thankyou very much for your explanation, I've pretty much got my head around it. Am I right i thinking the second line to the third all of the x change to x+y and is that a sort of rule when using summation? Thanks

 

[statdad]

I wouldn't say a rule, but a common bit of work. It's similar to making a substitution in a definite integral.

The original sum in line 2 is

<br /> \sum_{x=y}^\infty \frac{\lambda^x (0.9)^{x-y}}{(x-y)!}<br />

The form of the summand is similar to the infinite series for an exponential, but the starting value isn't zero. Suppose I use a new index
of summation, defined as

<br /> t = x - y \quad \text{ so } \quad x = t+y<br />

Since the original sum begins at x = y, the rewritten form begins at t = x - x = 0. In terms of the new variable the sum looks like

<br /> \sum_{t = 0}^\infty \frac{\lambda^{t+y} (0.9)^t}{t!} = \lambda^y \sum_{t=0}^\infty \frac{\lambda^t (0.9)^t}{t!}<br />

Writing this new form with summation index equal to x gives the form mentioned above.

 

[Homewoodm01]

Thanks for your time, you've really helped me, I'm confident I understand this now. Thanks again!