Calculate 5Σ r=0 r(r+1): Find the Sum!

[Nubcake]

Calculate 5 \Sigma r=0 r(r+1)
(Sorry I don't know how to do the proper notation online)

How do you calculate the sum for this since the common difference is changing?
I tried to write them out separately so the sum of r x sum of r+1 but I don't know how you put that in the sum formula.

Sn = n/2[2a + (n-1)d]
Sn = n/2[a + L]

 

[haruspex]

Hi Nubcake,
Given terms that grow like r², how fast do you think the sum will grow?

 

[Mark44]

Calculate 5 \Sigma r=0 r(r+1)
(Sorry I don't know how to do the proper notation online)

I don't know what you're asking.

How do you calculate the sum for this since the common difference is changing?
I tried to write them out separately so the sum of r x sum of r+1 but I don't know how you put that in the sum formula.

Sn = n/2[2a + (n-1)d]
Sn = n/2[a + L]

 

[SammyS]

Calculate 5 \Sigma r=0 r(r+1)
(Sorry I don't know how to do the proper notation online)

Do you mean \displaystyle \sum_{r=0}^{5}r(r+1)\ ?

 

[Curious3141]

Calculate 5 \Sigma r=0 r(r+1)
(Sorry I don't know how to do the proper notation online)

How do you calculate the sum for this since the common difference is changing?
I tried to write them out separately so the sum of r x sum of r+1 but I don't know how you put that in the sum formula.

Sn = n/2[2a + (n-1)d]
Sn = n/2[a + L]

Assuming you meant the sum as SammyS wrote it, there are at least a couple of ways to do it.

The most elementary way, but one that involves making a non-trivial observation, is to use what's called the "method of differences". Observe that (r+1)^3 - r^3 = 3r(r+1) + 1 What happens if you write out the sum of the left hand side? What terms cancel?

The less elementary way, but one that's easier to see if you know just a little calculus, is to let f(x) = x^{r+1}. Now differentiate twice, and work out the value at x = 1, i.e. calculate f''(1). What do you notice?

 

[Mark44]

The less elementary way, but one that's easier to see if you know just a little calculus, is to let f(x) = x^{r+1}. Now differentiate twice, and work out the value at x = 1, i.e. calculate f''(1). What do you notice?

Given that this was posted in the Precalc section, a solution involving calculus might not be applicable here.

 

[HallsofIvy]

The most obvious way to do \sum_{r=0}^5 r(r+1) is to write out all six terms and sum them: 0(0+1)+ 1(1+1)+ 2(2+1)+ 3(3+1)+ 4(4+1)+ 5(5+ 1)= 0+ 2+ 6+ 12+ 20+ 30= 70.

 

[Curious3141]

Given that this was posted in the Precalc section, a solution involving calculus might not be applicable here.

That does not mean the OP has no knowledge of it. Besides, it's better to suggest alternative solutions, especially if they're more direct and easier to see. The OP is free to disregard this solution if he so wishes.

 

[Mark44]

Given that this was posted in the Precalc section, a solution involving calculus might not be applicable here.

That does not mean the OP has no knowledge of it. Besides, it's better to suggest alternative solutions, especially if they're more direct and easier to see. The OP is free to disregard this solution if he so wishes.

I realize that problems are posted to the wrong sections all the time, and that is why I qualified what I said. On the other hand, the OP did post this in the precalc section, so I take that as a clue to the OP's current abilities. A calculus approach might seem more direct and easier to those of us with more knowledge, but would probably be completely mystifying to someone who isn't ready for such an approach yet.

 

[Curious3141]

I realize that problems are posted to the wrong sections all the time, and that is why I qualified what I said. On the other hand, the OP did post this in the precalc section, so I take that as a clue to the OP's current abilities. A calculus approach might seem more direct and easier to those of us with more knowledge, but would probably be completely mystifying to someone who isn't ready for such an approach yet.

I think it's pointless to keep going on about this. It's really for the OP to reply, choose a method and show more work.

 

[Mark44]

I agree completely.