# Power of intergers

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• October 5th 2009, 08:54 PM
flexus
Power of intergers
Find the sum, using the formulas for the sums of the powers of integers.

11
(-2-n^3+2n)
n=1
• October 5th 2009, 08:57 PM
Chris L T521
Quote:

Originally Posted by flexus
Find the sum, using the formulas for the sums of the powers of integers.

11
(-2-n^3+2n)
n=1

You need to know that:

$\sum_{i=1}^{n}k=kn$

$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

$\sum_{i=1}^{n}i^3=\left(\frac{n(n+1)}{2}\right)^2$

Can you tackle the problem now?
• October 5th 2009, 09:05 PM
thepride
Quote:

Originally Posted by Chris L T521
You need to know that:

$\sum_{i=1}^{n}k=kn$

$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

$\sum_{i=1}^{n}i^3=\left(\frac{n(n+1)}{2}\right)^2$

Can you tackle the problem now?

• October 5th 2009, 09:10 PM
flexus
i dont think that is the right answer the pride.. i see what you did though. you plugged 11 into n. you used the second equation. thanks for helping. the answer is will not help me if i don't understand this. can someone confirm his answer with a little explanation?
• October 5th 2009, 09:14 PM
Chris L T521
Quote:

Originally Posted by flexus
i dont think that is the right answer the pride.. i see what you did though. you plugged 11 into n. you used the second equation. thanks for helping. the answer is will not help me if i don't understand this. can someone confirm his answer with a little explanation?

Note that $\sum_{n=1}^{11}\left(2-n^3+2n\right)=\sum_{n=1}^{11}2-\sum_{n=1}^{11}n^3+2\sum_{n=1}^{11}n$.

Now substitute n=11 into the formulas I listed above and that will give you the answer you're looking for.
• October 5th 2009, 09:24 PM
flexus
Quote:

Originally Posted by Chris L T521
You need to know that:

$\sum_{i=1}^{n}k=kn$

$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

$\sum_{i=1}^{n}i^3=\left(\frac{n(n+1)}{2}\right)^2$

Can you tackle the problem now?

hey im looking at the first equation you gave me and i dont seem to know what to substitute k for?? so n=11 what does k =?
• October 5th 2009, 09:26 PM
Chris L T521
Quote:

Originally Posted by flexus
hey im looking at the first equation you gave me and i dont seem to know what to substitute k for?? so n=11 what does k =?

K is any constant. In your particular problem, k=2.
• October 5th 2009, 09:31 PM
flexus
Quote:

Originally Posted by Chris L T521
K is any constant. In your particular problem, k=2.

you mean -2 from the oringinal problem correct?

so if i set it up it would look like this?

(-2)(11)-11(11+1)/2+(11(11+1)^2/2= 704

correct?
• October 5th 2009, 09:39 PM
Chris L T521
Quote:

Originally Posted by flexus
you mean -2 from the oringinal problem correct?

so if i set it up it would look like this?

(-2)(11)-11(11+1)/2+(11(11+1))^2/4= 704 (change the value on the RHS and then you should have the answer)

correct?

Yes, I meant to say -2 -- my mistake.

You're so close -- note my corrections in red.
• October 5th 2009, 09:47 PM
flexus
Quote:

Originally Posted by Chris L T521
Yes, I meant to say -2 -- my mistake.

You're so close -- note my corrections in red.

omg i wrote the problem wrong. ok let me do this over.

the correct problem is:

15
(-2-n^3+2n)
n=1

ok so now i plug that into the three equations.

(-2)(15)-(15(15+1))/2+15(15+1)^2)/2 the i get:

-30-240/2+1920 which equals =1770 but this does not match the answer in my book. i must have done something wrong? can you confirm please?
• October 5th 2009, 09:50 PM
Chris L T521
Quote:

Originally Posted by flexus
omg i wrote the problem wrong. ok let me do this over.

the correct problem is:

15
(-2-n^3+2n)
n=1

ok so now i plug that into the three equations.

(-2)(15)-(15(15+1))/2+15(15+1)^2)/2 the i get:

-30-240/2+1920 which equals =1770 but this does not match the answer in my book. i must have done something wrong? can you confirm please?

It should be (-2)(15)- [(15(15+1))/2]^2+2[(15)(15+1)/2]...

Do you see why?
• October 5th 2009, 09:56 PM
flexus
Quote:

Originally Posted by Chris L T521
It should be (-2)(15)- [(15(15+1))/2]^2+2[(15)(15+1)/2]...

Do you see why?

so you are gonna distribute the negative to the right?
• October 5th 2009, 09:58 PM
Chris L T521
Quote:

Originally Posted by flexus
so you are gonna distribute the negative to the right?

For the second term? Yes.
• October 5th 2009, 10:04 PM
flexus
Quote:

Originally Posted by Chris L T521
For the second term? Yes.

ok im gonna try this again. my head is about to explode lol

(-2)(15)-[(15(15+1))/2]^2 + 2[(15)(15+1)/2]

= -30 -14400+14400

=-30

OK that's what i got. can you confirm please.
• October 5th 2009, 10:41 PM
Chris L T521
Quote:

Originally Posted by flexus
ok im gonna try this again. my head is about to explode lol

(-2)(15)-[(15(15+1))/2]^2 + 2[(15)(15+1)/2]

= -30 -14400+14400

=-30

OK that's what i got. can you confirm please.

The term in red should be 240 (since its not being squared).

You should end up with -14190
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