# Maths summation algebra

1. Jan 3, 2010

### Homewoodm01

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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

Last edited by a moderator: May 4, 2017
2. Jan 3, 2010

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

$$\binom x y= \frac{x!}{y! (x-y)!}$$

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

$$\sum_{x=0}^\infty \lambda^{x+y} \frac{0.9^x}{x!} = \lambda^y \sum_{x=0}^\infty \frac{(0.9\, \lambda)^x}{x!}$$

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

3. Jan 3, 2010

### 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

4. Jan 3, 2010

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

$$\sum_{x=y}^\infty \frac{\lambda^x (0.9)^{x-y}}{(x-y)!}$$

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

$$t = x - y \quad \text{ so } \quad x = t+y$$

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

$$\sum_{t = 0}^\infty \frac{\lambda^{t+y} (0.9)^t}{t!} = \lambda^y \sum_{t=0}^\infty \frac{\lambda^t (0.9)^t}{t!}$$

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

Last edited: Jan 3, 2010
5. Jan 3, 2010

### Homewoodm01

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