proof by induction help

Using standard result for∑_(r=1)^n▒r^3
Show that ∑_(r=n+1)^2n▒r^3 = n2/4(3n+1)(5n+3)
I began with n=2
2=2
Assume n=k
=∑_(r=k+1)^2k▒r^3 = k2/4(3k+1)(5k+3)
Not sure of the next step because of the 2n?

Thanks in advance :smile:

Reply 1

Notnek

21

So you need to show that it holds for n=k+1 --> 2n=2k+2. Next notice that

\displaystyle \sum_{r=k+2}^{2k+2}r^3 = \sum_{r=k+1}^{2k}r^3 - \sum_{r=k+1}^{k+1}r^3 + \sum_{r=2k+1}^{2k+1}r^3 + \sum_{r=2k+2}^{2k+2}r^3

Does this help?

Reply 2

ian.slater

12

Original post by Petro_99

Using standard result for∑_(r=1)^n▒r^3
Show that ∑_(r=n+1)^2n▒r^3 = n2/4(3n+1)(5n+3)
I began with n=2
2=2
Assume n=k
=∑_(r=k+1)^2k▒r^3 = k2/4(3k+1)(5k+3)
Not sure of the next step because of the 2n?

Thanks in advance :smile:

I think the question is not asking you to use induction but to use the standard formula for sum of cubes a couple of time to work out the sum between the limits they ask for.

Reply 3

Notnek

21

Original post by ian.slater

I think the question is not asking you to use induction but to use the standard formula for sum of cubes a couple of time to work out the sum between the limits they ask for.

Having looked at the question again, I agree with you.

OP

so what do i do?

sum r = half n(n+1)

ring any bells?

sub all that in then typically there's some rearranging to do to match the format

Reply 6

ian.slater

12

Original post by Petro_99

so what do i do?

You could start by working out the sum of the first 2n cubes. You need the standard formula for summing cubes which should be in your book. But you need to sum to 2n, not just to n.

Reply 7

Petro_99

OP

10

Original post by mucgoo

sum r = half n(n+1)

ring any bells?

sub all that in then typically there's some rearranging to do to match the format

Original post by ian.slater

You could start by working out the sum of the first 2n cubes. You need the standard formula for summing cubes which should be in your book. But you need to sum to 2n, not just to n.

ahhh! lol

thanks for your help! :biggrin:

proof by induction help

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