1. ## Partial Sum/Sequence questions...

Obviously I'm making some sort of systematic error...any help is greatly appreciated!

15.) The number of cars sold weekly by a new auto dealership grows according to a linear growth model. The first week the dealership sold two cars $\displaystyle P_0=2$ and the second week the dealer ship sold six cars $\displaystyle P_1=6$

In the first 50 weeks, how many cars did the dealership sell?

My work:

$\displaystyle S_n=\frac{(P_0+P_{n-1})(n)}{2}$

$\displaystyle P_{n-1}=P_{49-1}=P_{48}$

$\displaystyle P_{48}=P_0+nd=2+48(4)=194$

$\displaystyle S_{49}=\frac{(2+194)(49)}{2}$

$\displaystyle S_{49}=4802$

17.) $\displaystyle 5+8+11+14+...+299+302=?$ (100 terms)

My work:

$\displaystyle S_{99}=\frac{(5+299)(99)}{2}=15,048$

19.) $\displaystyle 15+11+7+3+...$

Find the sum of the first 100 terms

My work:

$\displaystyle P_{98}=15+(-4)(98)=-377$

$\displaystyle S_{99}=\frac{(P_0+P_{n-1})(n)}{2}=\frac{(15+(-377))(99)}{2}=-17,919$

Like I said, I'm pretty sure it's a systematic error I'm making...I've tried it lots more way besides this, but I think this is the "most" correct way I've tried....any help will be wonderful. Thanks!!

2. Originally Posted by elizsimca

17.) $\displaystyle 5+8+11+14+...+299+302=?$ (100 terms)

\displaystyle \begin{aligned}\underbrace{5+8+11+\cdots}_{100\tex t{ terms}}&=\sum_{n=0}^{99}\left\{5+3n\right\}\\ &=5(99+1)+3\cdot\frac{99\cdot{100}}{2}\\ &=15350\end{aligned}

Using $\displaystyle \sum_{n=0}^{N}c=c(N+1)$

and $\displaystyle \sum_{n=0}^{N}n=\frac{N(N+1)}{2}$

And
19.) $\displaystyle 15+11+7+4+\cdots=?$

100 terms

Find the sum of the first 100 terms
\displaystyle \begin{aligned}\underbrace{15+11+7+4+\cdots}_{100\ text{ terms}}&=\sum_{n=0}^{99}\left\{15-4n\right\}\\ &=15(99+1)-4\cdot\frac{99\cdot100}{2}\\ &=-18300\end{aligned}

3. Originally Posted by Mathstud28
\displaystyle \begin{aligned}\underbrace{5+8+11+\cdots}_{100\tex t{ terms}}&=\sum_{n=0}^{99}\left\{5+3n\right\}\\ &=5(99+1)+3\cdot\frac{99\cdot{100}}{2}\\ &=15350\end{aligned}

Using $\displaystyle \sum_{n=0}^{N}c=c(N+1)$

and $\displaystyle \sum_{n=0}^{N}n=\frac{N(N+1)}{2}$

And

\displaystyle \underbrace{15+11+7+4+\cdots}_{100\text{ terms}}&=\sum_{n=0}^{99}\left\{15-4n\right\}\\ &=15(99+1)-4\cdot\frac{99\cdot100}{2}\\ &=-18300\end{aligned}
Mathstud,

So am I using an incorrect formula? Is the formula I tried to use on number 17 completely incorrect? That's the formula my professor gave us and I'm wondering if it was typed incorrectly because I'm getting all my partial arithmetic sums incorrect when I use that formula...when I do it your way, I get the correct answer....any thoughts?

Thanks

4. Originally Posted by elizsimca
Mathstud,

So am I using an incorrect formula? Is the formula I tried to use on number 17 completely incorrect? That's the formula my professor gave us and I'm wondering if it was typed incorrectly because I'm getting all my partial arithmetic sums incorrect when I use that formula...when I do it your way, I get the correct answer....any thoughts?

Thanks
No! You are in fact using the right formula ...but unfortunately you are putting in the wrong numbers

$\displaystyle S_{100}=\frac{\left(5+{\color{red}302}\right)\cdot {\color{red}100}}{2}=15350$

Dont worry those upper indexes always get people

5. Originally Posted by elizsimca
15.) The number of cars sold weekly by a new auto dealership grows according to a linear growth model. The first week the dealership sold two cars $\displaystyle P_0=2$ and the second week the dealer ship sold six cars $\displaystyle P_1=6$

In the first 50 weeks, how many cars did the dealership sell?

My work:

$\displaystyle S_n=\frac{(P_0+P_{n-1})(n)}{2}$

$\displaystyle P_{n-1}=P_{49-1}=P_{48}$

$\displaystyle P_{48}=P_0+nd=2+48(4)=194$

$\displaystyle S_{49}=\frac{(2+194)(49)}{2}$

$\displaystyle S_{49}=4802$

You are taking $\displaystyle n=50$ so you should be working out $\displaystyle S_{50}=\frac{(P_0+P_{49})(50)}2.$

6. Originally Posted by Mathstud28
No! You are in fact using the right formula ...but unfortunately you are putting in the wrong numbers

$\displaystyle S_{99}=\frac{\left(5+{\color{red}302}\right)\cdot{ \color{red}100}}{2}=15350$

Dont worry those upper indexes always get people

But if $\displaystyle P_0=$ Week 1...then wouldn't the 50th week be given by $\displaystyle P_{49}$?

Sorry, I'm having a dumb math moment....lol

7. Originally Posted by elizsimca
But if $\displaystyle P_0=$ Week 1...then wouldn't the 50th week be given by $\displaystyle P_{49}$?

Sorry, I'm having a dumb math moment....lol
But how many terms are there?

8. Originally Posted by Mathstud28
But how many terms are there?
touche'