Summation

Hello I know that the summation of r^2 from r=1 to r=n is [n(n+1)(2n+1)]/6, but I wasn't able to derive it. My approach was to write it as 1,1+2+1,1+4+4,1+6+9... but it didn't get me anywhere. A bit of aid would be much appreciated.

Reply 1

TeeEm

19

Original post by Spandy

Hello I know that the summation of r^2 from r=1 to r=n is [n(n+1)(2n+1)]/6, but I wasn't able to derive it. My approach was to write it as 1,1+2+1,1+4+4,1+6+9... but it didn't get me anywhere. A bit of aid would be much appreciated.

there are several different proofs

e.g induction

or

method of differences with a choice of suitable identities

Reply 2

Spandy

OP

3

Oh crap, induction had totally slipped out of my mind, thanks!!

Reply 3

TeeEm

19

Original post by Spandy

Oh crap, induction had totally slipped out of my mind, thanks!!

my pleasure

Reply 4

NoNewFriends

Do you know how to derive Σk? Whatever method you used for this also applies here!

Reply 5

TenOfThem

18

Original post by NoNewFriends

Do you know how to derive Σk? Whatever method you used for this also applies here!

No it doesn't

Unless I have completely misunderstood what you are suggesting

Reply 6

NoNewFriends

Original post by TenOfThem

No it doesn't

Unless I have completely misunderstood what you are suggesting

You know a lot more than I know so I'm probably wrong but which method do you have in mind that works for Σk but doesn't for Σk²?

Reply 7

TenOfThem

18

Original post by NoNewFriends

You know a lot more than I know so I'm probably wrong but which method do you have in mind that works for Σk but doesn't for Σk²?

Standard proof for the sum of the first n numbers

Reply 8

NoNewFriends

Original post by TenOfThem

Standard proof for the sum of the first n numbers

Which one is the standard one?

Reply 9

TenOfThem

18

Original post by NoNewFriends

Which one is the standard one?

Write the numbers in reverse
Add to the original series
Half

Reply 10

joostan

13

Original post by NoNewFriends

Which one is the standard one?

Let:

S_n= \displaystyle\sum_{k=1}^n k

Then

S_n=1+2+. . .+(n-1)+n[br]\Rightarrow S_n=n+(n-1)+. . .+2 +1[br]\Rightarrow 2S_n=\underbrace{(n+1)+(n+1)+. . .+(n+1)+(n+1)}_{\text{n \ times}}[br]\Rightarrow S_n=\dfrac{n(n+1)}{2}

Reply 11

Spandy

OP

3

Original post by NoNewFriends

Which one is the standard one?

Hurrah for Joostan. Or simply use the formula for arithmetic progressions

Reply 12

WhiteGroupMaths

9

I wrote this a long time back on my website (http://www.whitegroupmaths.com/2010/04/series-sequences-and-mathematical.html see Q6), hopefully it helps. Peace.

Reply 13

WhiteGroupMaths

9

I wrote this a long time back on my website (http://www.whitegroupmaths.com/2010/04/series-sequences-and-mathematical.html see Q6), hopefully it helps. Peace.

(Click on the image to enlarge)

Reply 14

TenOfThem

18

Original post by Spandy

Hurrah for Joostan. Or simply use the formula for arithmetic progressions

You were looking for proof

Reply 15

Spandy

OP

3

Original post by TenOfThem

You were looking for proof

Uhh, I must be drunk.
My bad...

Reply 16

DFranklin

18

Original post by WhiteGroupMaths

I wrote this a long time back on my website (http://www.whitegroupmaths.com/2010/04/series-sequences-and-mathematical.html see Q6), hopefully it helps. Peace.

Or similarly but with r(r+1)(r+2) - (r-1)r(r+1) = 3r(r+1).

Summation

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