/sci/ - Science & Math

[Anonymous No.1499375]

/sci/ can't into induction

show that <span class="math">1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2[/spoiler]

nb4 homework

[Anonymous No.1499383]

wat

[Anonymous No.1499408]

i'll do induction start: n=1:
1^3 = 1^2

you guys do the rest.

[Anonymous No.1499423]

how do I n=k, n=(k+1) ???

[Anonymous No.1499430]

wat

[Anonymous No.1499522]

you fucking fail /sci/. i never want to talk to you again.

not even the spambot got the right answer.

[Anonymous No.1499538]

NEEEEEEERDSSSS

[Anonymous No.1499593]

trickier than it looks

[Anonymous No.1499598]

because God.

[Anonymous No.1499626]

(1 + 2 + ... + n)^2 = (n(n+1)/2)^2 = 1^3 + ... + n^3

Is induction really necessary? Induction seems like a more painful way to go about this.

[Anonymous No.1499634]

ok kids, >>1499408
did the base case.
Let's start with the strong induction. We'll assume that 1^3 + 2^3 + ...+ k^3 = (1+2+...+k)^2 for some integer k >0

now, finish it

[Anonymous No.1499665]

>>1499626
if you use the formula in the middle (gauss formula) squared and add (n+1)^3 to it, it should algebraically work out to be ((n+1)(n+2)/2)^2.
I don't think that's very hard, and it's induction.

[Anonymous No.1499676]

>>1499626
while your equation is true, you have yet to prove anything. If you can reduce the series on the rhs to a handy dandy equation, then you'll be done

[Anonymous No.1499683]

>>1499665

That was my thought too, but if you're going to use n(n+1)/2 why the hell not just use the fact that 1^3 + ... + n^3 = (n(n+1)/2)^2 and one-line it instead of induction?

Well, it's OP's homework I guess.

[Anonymous No.1499680]

so /sci/ im looking into colleges to apply to and was wondering if y'all could help.

i live in michigan and am looking into a biochem/biotech major. i got a 30 on my ACT and have an adjusted GPA somewhere in the 4's from a good HS. is UoM my best bet or ist here soempalce significantly better in my general region for such a major? pic unrelated

[Anonymous No.1499695]

>>1499665
this.
at least this post is actually about math

[Anonymous No.1499697]

Results.

[Anonymous No.1499700]

OP, we have that <span class="math">\sum_{i=1}^ni=\frac{n(n+1)}{2}[/spoiler]. So square that, add <span class="math">(n+1)^3[/spoiler] to it and you get <span class="math">\frac{(n+1)(n+2)}{2}[/spoiler], which is <span class="math">(1+2+\cdotd+(n+1))^2[/spoiler].

[Anonymous No.1499703]

>>1499683
because I didn't know that formula?
I mean, how do you think that closed form is proven? Without induction? You wish. (Well, perhaps there are in fact some crazy combinatorial maniacs out there who do it without induction. The fools.)

btw, I'm OP, it's not homework. It's just a nice identity I found by coincidence.

[Anonymous No.1499708]

>>1499700
typo, the unknown sequence is <span class="math">(1+2+\cdots+(n+1))^2[/spoiler].

[Anonymous No.1499709]

HEY CHRISTAN
If yuor story is real why don't you act in reckless endangerment to get to heaven faster?

Cant kill yourself but if you walk across a plank 90ft above the ground blindfolded you can still get into heaven

[None !1Z.c8wZLWc No.1499718]

>>1499634

for n = 1

1^3 = 1^2 ok rite

now for k+1 it'd be

1^3 + 2^3 + ... + k^3 + (k+1)^3 = (1 + 2 + .... + k + k +1)^2

since we assumed what we assumed...

1^3 + 2^3 + ... + k^3 + (k+1)^3 = (1 + 2 + ... + k)^2 + (k+1)^3 = \ hmmm u gotta do some transformations here and if the original equation is true u'll get \ = (1 + 2 + .... + k + k +1)^2

[Anonymous No.1499726]

>>1499634

1^3 + 2^3 +... k^3 + (k+1)^3 = (1 + 2 + .... + k + (k+1))^2 = ((k+1)(k+2)/2)^2 = [(k+1)^2 * (k+2)^2]/4 = [{(k^2+ 2k + 1)*(k^2 + 4k + 4)}/4]
= [(k^4 + 6k^3 + 13k^2 + 12k + 4)/4]
= [{(k^2)*(k^2 + 2k + 1) + 4k^3 + 12k^2 + 12k +4}/4]
= [{(k^2)*(k+1)^2}/4] + [4k^3 +12k^2 + 12k +4]/4
= [{k*(k+1)}/2]^2 + k^3 + 3k^2 + 3k + 1
= [{k*(k+1)}/2]^2 + (k+1)^3

Q.E.D.

[Anonymous No.1499746]

You don't need a PhD to understand 7th grade math

[Anonymous No.1499747]

1*3+2^3+...+n^3 = (1+2+...+n)^2

Check for n = 1

1^3 = 1 = (1)^2

Assume proof now works for some arbitrary value "n"

Now check to see if it works for "n = n +1"

[1^3 +2^3 +...+n^3] + (n+1)^3 = (1+2+...+n+n+1)^2

Square bracket can be rewritten as (1+2+...+n)^2

(1+2+...+n)^2 + (n+1)^3 = [(1+2+...n+(n+1))^2]

Square brackets can be rewritten as (1+2+...+n)^2 + 2(n)*(n+1)+(n+1)^2

Subtract (1+2+...+n)^2 from both sides

(n+1)^3 = 2(n)*(n+1) + (n+1)^2

Left sides expands to n^3 + 2n^2 + 2n+ 1

Right side expands to 2n^2 + 2n + n^2 + 2n + 1 =........ 3n^2 + 4n + 1

???

Well fuck, I thought I had it, I give.

[Anonymous No.1499757]

>>1499703

You definitely don't need to use induction to prove those.

1 + 2 + ... + n = n + ... + 2 + 1
Pair up the terms and you get (n+1) + (n+1) + ... n times, so n(n+1).
Divide by 2.

Or you can be like derp 1 + 2 + ... + n = (n+1) choose 2 by hockey stick done.

For the other one, I think there's a combinatorial way but I can't remember it. You can get to it by summing (k+1)^4 - k^4 and telescoping and stuff.

[Anonymous No.1499774]

>>1499757

i think he meant

1^3 + 2^3 + ... + n^3 = [n*(n+1)/2]^2

[Anonymous No.1499784]

>>1499726
*FF victory theme*
looks like we have a winner. The rest of you can go home now

[Anonymous No.1499822]

>implying <span class="math">(1 + 2 + ... + n)^2 + (n+1)^3 = (1 + 2 + ... + n + (n+1) )^2 \forall n>3[/spoiler]