1. ## Power of intergers

Find the sum, using the formulas for the sums of the powers of integers.

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

2. 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?

3. 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?

4. 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?

5. 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.

6. 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 =?

7. 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.

8. 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?

9. 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.

10. 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?

11. 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?

12. 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?

13. Originally Posted by flexus
so you are gonna distribute the negative to the right?
For the second term? Yes.

14. 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.

15. 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

Page 1 of 2 12 Last