1. ## couple of questions...

I have a couple of questions that have got me bugged for a while...

1)Find the sum of the arithmetic progression 1, 4, 7, 10, 13, 16... , 1000.
(already figured that part out to be 167167, with 334 terms)

Every third term of the progression above is removed, i. e. 7, 16, etc. Find The sum of the remaining terms.

I figured that I'd just find the sum of the terms removed, then substract that from 167167, except for the fact that I don't know the number of terms are in the sequence. Any ideas?

2)The binomial expansion of (1+ax)^n, where n is a positive integer, has six terms. The coefficent of the x^3 term is 5/4. Find a.

I know that n=5, but the rest is a mystery to me algebraically. I did it using Pascal's Triangle, but I want to know how to do it a quicker way.

2. Originally Posted by DemonX01
2)The binomial expansion of (1+ax)^n, where n is a positive integer, has six terms. The coefficent of the x^3 term is 5/4. Find a.

I know that n=5, but the rest is a mystery to me algebraically. I did it using Pascal's Triangle, but I want to know how to do it a quicker way.
$(1 + ax)^5 = \sum_{r = 0}^5{5\choose{r}}(1)^{5 - r}(ax)^r$

$= \sum_{r = 0}^5{5\choose{r}}a^rx^r$.

The $x^3$ term will be of the form

${5\choose{3}}a^3x^3$

so its coefficient is

${5\choose{3}}a^3$.

You should know that ${5\choose{3}} = 10$.

And since this coefficient is $\frac{5}{4}$, then you have

$10a^3 = \frac{5}{4}$

$a^3 = \frac{1}{8}$

$a = \frac{1}{2}$.

3. Originally Posted by DemonX01
I have a couple of questions that have got me bugged for a while...

1)Find the sum of the arithmetic progression 1, 4, 7, 10, 13, 16... , 1000.
(already figured that part out to be 167167, with 334 terms)

Every third term of the progression above is removed, i. e. 7, 16, etc. Find The sum of the remaining terms.

I figured that I'd just find the sum of the terms removed, then substract that from 167167, except for the fact that I don't know the number of terms are in the sequence. Any ideas?
You should have written down a couple more steps: 7, 16, 25, 34,..., 997 is an arithmetic sequence with common difference 9 (because every third step is 3+ 3+ 3= 9). There are, of course, 333/3= 111 terms in that sequence.

2)The binomial expansion of (1+ax)^n, where n is a positive integer, has six terms. The coefficent of the x^3 term is 5/4. Find a.

I know that n=5, but the rest is a mystery to me algebraically. I did it using Pascal's Triangle, but I want to know how to do it a quicker way.[/QUOTE]

4. Originally Posted by HallsofIvy
You should have written down a couple more steps: 7, 16, 25, 34,..., 997 is an arithmetic sequence with common difference 9 (because every third step is 3+ 3+ 3= 9). There are, of course, 333/3= 111 terms in that sequence.
[/quote]

Could you explain that part more? How did you arrive at 997 as being the final term?