Partial Sum/Sequence questions...

**elizsimca** · November 30th 2008, 05:07 PM

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 $P_0=2$ and the second week the dealer ship sold six cars $P_1=6$

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

My work:

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

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

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

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

$S_{49}=4802$

The correct answer is 5000...

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

My work:

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

The correct answer is 15,350

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

Find the sum of the first 100 terms

My work:

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

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

Correct Answer: -18,300

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!!

**Mathstud28** · November 30th 2008, 05:18 PM

Originally Posted by elizsimca

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

The correct answer is 15,350

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

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

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

And

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

100 terms

Find the sum of the first 100 terms

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

**elizsimca** · November 30th 2008, 05:22 PM

Originally Posted by Mathstud28

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

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

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

And

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

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

**Mathstud28** · November 30th 2008, 05:26 PM

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

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

Dont worry those upper indexes always get people

**JaneBennet** · November 30th 2008, 05:27 PM

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 $P_0=2$ and the second week the dealer ship sold six cars $P_1=6$

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

My work:

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

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

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

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

$S_{49}=4802$

The correct answer is 5000...

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

**elizsimca** · November 30th 2008, 05:29 PM

Originally Posted by Mathstud28

No! You are in fact using the right formula

...but unfortunately you are putting in the wrong numbers

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

Dont worry those upper indexes always get people

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

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

**Mathstud28** · November 30th 2008, 05:30 PM

Originally Posted by elizsimca

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

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

But how many terms are there?

**elizsimca** · November 30th 2008, 05:32 PM

Originally Posted by Mathstud28

But how many terms are there?

touche'

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