Induction Q (1 Viewer)

[SSejychan]

Please help with this induction question:

Pn: 1^2 + 2^2 + 3^2 ...... + (2n + 1)^2 = [(n +1)(2n + 1)(2n + 3)]/6, n > 0

I can't even prove that P1 is true, let alone do the rest of the question...

here's my working:

P1:
LHS = 1^2 + [2(1) + 1]^2
= 1 + 3^2
= 10

RHS = [(1 +1)(2(1) + 1)(2(1) + 3)]/6
= [(2)(3)(5)]/6
= 5

what's wrong with this??? How do i prove P1 is true?? What am i doing wrong?

but, if i leave that for a while, and move on with the question:

Assume Pn true for n=k

Pk: 1^2 + 2^2 + 3^2 ...... + (2k + 1)^2 = [(k +1)(2k + 1)(2k + 3)]/6

Consider P(k+1)

LHS = 1^2 + 2^2 + 3^2 ...... + (2k + 1)^2 + (2(k + 1) + 1)^2
= [(k +1)(2k + 1)(2k + 3)]/6 + (2(k + 1) + 1)^2 {by assumption}
= [(k +1)(2k + 1)(2k + 3)]/6 + (2k + 3)^2
= [(k +1)(2k + 1)(2k + 3) + 6(2k + 3)^2]/6
= {[2k + 3][(k +1)(2k + 1) + 6(2k + 3)]}/6
= [(2k + 3)(2k^2 + 3k + 1 + 12k + 18)]/6
= [(2k + 3)(2k^2 + 15k + 19)]/6

RHS = [(k + 1 +1)(2(k + 1) + 1)(2(k + 1) + 3)]/6
= [(k +2)(2k + 3)(2k + 5)]/6
= [(2k + 3)(2k^2 + 9k + 10)]/6

But, LHS doesn't equal RHS!!

Could someone please tell me what's wrong with my working... and, if there's a better way to do this question.

 

[ssglain]

The first thing I noticed is that there seems to be something wrong with the LHS series.

Σ(2r + 1)² [from r = 0 -> r = n] should be 1² + 3² + 5² + ... + (2n + 1)².

Judging by the first three terms shown, the series is actually Σn² [from r = 0 -> r = n] and the last term should be n². Was that a typo?

But either way, it's not true for n = 1, which leads me to question whether the range of n is correct.

Where is this question from?

 

[SSejychan]

The first thing I noticed is that there seems to be something wrong with the LHS series.

Σ(2r + 1)² [from r = 0 -> r = n] should be 1² + 3² + 5² + ... + (2n + 1)².

Judging by the first three terms shown, the series is actually Σn² [from r = 0 -> r = n] and the last term should be n². Was that a typo?

But either way, it's not true for n = 1, which leads me to question whether the range of n is correct.

Where is this question from?

i got the question from this math resources site: http://www.geocities.com/mathematicsplus/resources.html

and, here's the link to the actual question (it's the first one): http://www.geocities.com/mathematicsplus/y11_pre_maths_ext1_worksheet_harder_induction.pdf

edit: yeah, the sheet says it's from the preliminary course, but i doubt we did binomial theorem and complex numbers in yr 11...

and, yeah.. i was really confused, because if it is the sum of (2n+1)^2, then the 1^2 doesn't fit into it, for n>0....

 

[SSejychan]

Hey, you're right. =) It works if the denominator is 3!

Thanks guys for helping.

edit: huh??? where did Trebla's post go...

well. thanks Trebla for pointing that discrepancy out. =)

ok, here's the correct answer:

Pn: 1^2 + 2^2 + 3^2 ...... + (2n + 1)^2 = [(n +1)(2n + 1)(2n + 3)]/3, n > 0

P1:
LHS = 1^2 + [2(1) + 1]^2
= 1 + 3^2
= 10

RHS = [(1 +1)(2(1) + 1)(2(1) + 3)]/3
= [(2)(3)(5)]/3
= 10

Therefore, P1 is true

Assume Pn true for n=k

Pk: 1^2 + 2^2 + 3^2 ...... + (2k + 1)^2 = [(k +1)(2k + 1)(2k + 3)]/3

Consider P(k+1)

LHS = 1^2 + 2^2 + 3^2 ...... + (2k + 1)^2 + (2(k + 1) + 1)^2
= [(k +1)(2k + 1)(2k + 3)]/3 + (2(k + 1) + 1)^2 {by assumption}
= [(k +1)(2k + 1)(2k + 3)]/3 + (2k + 3)^2
= [(k +1)(2k + 1)(2k + 3) + 3(2k + 3)^2]/3
= {[2k + 3][(k +1)(2k + 1) + 3(2k + 3)]}/3
= [(2k + 3)(2k^2 + 3k + 1 + 6k + 9)]/3
= [(2k + 3)(2k^2 + 9k + 10)]/3

RHS = [(k + 1 +1)(2(k + 1) + 1)(2(k + 1) + 3)]/3
= [(k +2)(2k + 3)(2k + 5)]/3
= [(2k + 3)(2k^2 + 9k + 10)]/3

LHS = RHS

Hence, etc etc...

 

[ssglain]

I still think it's a dodgy question. The nth term of the LHS series is clearly defined by T(n) = (2n + 1)², but 2² doesn't fit in anywhere if n is an integer...

 

[SSejychan]

oh.... crap...

sorry, guys... yes, it WAS a typo, and i overlooked it completely..

SORRY.

It's actually.

Pn: 1^2 + 3^2 + 5^2 ...... + (2n + 1)^2 = [(n +1)(2n + 1)(2n + 3)]/3, n > 0
so so so sorry!!!!!

 

[ZJ NingNing]

oh.... crap...

sorry, guys... yes, it WAS a typo, and i overlooked it completely..

SORRY.

It's actually.

Pn: 1^2 + 3^2 + 5^2 ...... + (2n + 1)^2 = [(n +1)(2n + 1)(2n + 3)]/3, n > 0
so so so sorry!!!!!

LOL!

what was funny was that in the email you sent me about this question, the one that you meant to correct your typo, you did another one. So I was damn confused.

and I did try the question with a denominator of 3, but it didnt work either cos of your damn typo.

way too many 'sorry's from you, my dear chan.

PS I wont be on MSN today or tomorrow I think. But next week i should be on there more.

 

[ssglain]

oh.... crap...

sorry, guys... yes, it WAS a typo, and i overlooked it completely..

SORRY.

It's actually.

Pn: 1^2 + 3^2 + 5^2 ...... + (2n + 1)^2 = [(n +1)(2n + 1)(2n + 3)]/3, n > 0
so so so sorry!!!!!

Hahaha. I thought that was the case.

 

[sen00]

prove this:

[(n+1)(2n+1)]^n >= (n!)^2*6^n