It's not obvious. It would be either +6 or +6b when you evaluate the sum.
Looking at the numbers, it Is guess +6b may make more sense, but will check now.

(edited 3 years ago)

Reply 5

ghostwalker

Once expanded, +6 rather than +6b works nicely.

Reply 6

the bear

Original post by ghostwalker

Once expanded, +6 rather than +6b works nicely.

Reply 7

ghostwalker

Original post by the bear

lol

Reply 8

beachpanda

I've taken the 6 as outside the sum and now have this - but still unsure how to solve when I've polynomials in both variables?

0-14.jpg541.5KB

Reply 9

mqb2766

What are the surrounding questions like?
They may just want you to sub a few values in?

Reply 10

beachpanda

Original post by mqb2766

What are the surrounding questions like?
They may just want you to sub a few values in?

Other questions have just been about substituting for the standard results where the sigma notation is - I'm about to learn method of differences?

(edited 3 years ago)

Reply 11

mqb2766

Original post by beachpanda

Other questions have just been about substituting for the standard results where the sigma notation is - pre-method of differences?

The answer is a small number(s) which can be obtained by writing the first few values down. That may be sufficient?
Understanding a) and c) is not too hard as c) is the square of triangular numbers (a) with different index which you can do a bit of reasoning about. Possibly beyond what is required.

(edited 3 years ago)

Reply 12

beachpanda

Original post by ghostwalker

Once expanded, +6 rather than +6b works nicely.

Just revisiting this - I've done the attached to try and solve a & b but not sure how I can solve this?

Foto am 04-03-2021 um 12.07 #2.jpg68.5KB

Reply 13

ghostwalker

Original post by beachpanda

Just revisiting this - I've done the attached to try and solve a & b but not sure how I can solve this?

Simplest way is just to write down the first few partial sums for each summation and look for the common term as mqb2766 suggested.

Reply 14

beachpanda

Original post by ghostwalker

Simplest way is just to write down the first few partial sums for each summation and look for the common term as mqb2766 suggested.

Is this what you mean? Just checked the answers and a ≠ c

Foto am 04-03-2021 um 12.31.jpg41.2KB

Reply 15

mqb2766

You'd need a few more "a" terms, triangular numbers, to get a match with b.

Reply 16

DFranklin

FWIW, since the sum of cubes is a square, it's relatively easy to evaluate the middle sum until you get to squares which narrows things down quite quickly.

Reply 17

beachpanda

Original post by mqb2766

You'd need a few more "a" terms, triangular numbers, to get a match with b.

Does this look right? The sums in a and b were both equal for 36, so I subbed the value of r in each sum - then equated

Still doesn't look right considering a = 8, b = 4 from the answer section

Foto am 04-03-2021 um 13.04.jpg67.6KB

Reply 18

mqb2766

Original post by beachpanda

Does this look right? The sums in a and b were both equal for 36, so I subbed the value of r in each sum - then equated

Still doesn't look right considering a = 8, b = 4 from the answer section

The 8th triangular number is 36, as is 1+4+9+16+6. So a=8, b=4.

Reply 19

mqb2766

Original post by beachpanda

Does this look right? The sums in a and b were both equal for 36, so I subbed the value of r in each sum - then equated

Still doesn't look right considering a = 8, b = 4 from the answer section

I guess the question is less about doing algebra in two variables, more about interpreting the sums. The third sum is simply the square of the values in a, or the sum of cubes is the triangular numbers squared.