Question

Prove by induction for n>= 1
1(2) + 2
(3) + 3
(4) +...+ n(n + 1) =
(n(n + 1)(n + 2))/3

Answer 1

step 1: prove it first for n=1

Answer 2

have you done that?

Answer 3

yes, I'm failing on the inductive step
also proved it for n=2, n=3 for kicks

Answer 4

ok, let me try it with you and we'll see how it goes...

Answer 5

tried 1(2) + 2(3) + 3(4) + n(n + 1) + (n(n + 1))(n(n + 2)) =, does the left side seem right?

Answer 6

so that is where your mistake probably lies...

Answer 7

\[n=1:\]\[1(2)={1(1+1)(1+2)\over3}=2\]\[n:\]\[1(2) + 2(3) + 3(4) +...+ n(n + 1) ={
(n(n + 1)(n + 2))\over3}\]\[n+1:\]\[1(2)+2(3)+...+n(n+1)+(n+1)(n+2)\]\[={
(n(n + 1)(n + 2))\over3}+(n+1)(n+2)\]hope the algebra works out...

Answer 8

got a math processing error, i'm guessing the error is in the +2 section

Answer 9

you can't read above?

Answer 10

came through error

Answer 11

1(2) + 2(3) + 3(4) + n(n + 1) + (n+1)(n+2) = ?does the left side look right now?

Answer 12

well I did
1(2)+2(3)+...+n(n+1)+(n+1)(n+2)
yeah there we go

Answer 13

cool, but the right side will end up cubic, and the left side quadratic

Answer 14

you don't need to worry about the left, we just need to get
((n+1)(n + 2)(n + 3))/3
on the right and we will have proven it for n+1, and therefor all n

Answer 15

summation of n is n(n+1)/2
summation of n^2 is n(n+1)(2n+1)/6
adding these we get, your required result

Answer 16

@experimentX that is not the problem we are doing

Answer 17

but don't we want the left side to equal the right side, and will the proof on the right side be (n(n + 1)(n + 2))/3 ?

Answer 18

somewhat like this work??
http://www.wolframalpha.com/input/?i=sum+j%2Bj%5E2%2C+j%3D1+to+n

Answer 19

no, we have to show that if we add one more term to the left, then the right side will give the same formula, but with n+1 in place of n

Answer 20

so we don't even work with the right, just show that the right becomes one more than our hypothesis

Answer 21

the algebra ends up being 2n^2 + 4n + 22 for n = 4, but now what?

Answer 22

we are proving that the truth of the statement for n \(implies\) that it is true for n+1
sorry, but I don't think you understand the concept here jobu

Answer 23

@TuringTest i mean it is same as the summation of j+j^2 for 1 to n, isn't it

Answer 24

yes, turing, you understood it, but I'm stuck with what to do with the algebra after calculating the N+1

Answer 25

we don't need to try various numbers for n
@experimentX no it would be\[\sum_{i=1}^{n}i(i+1)\]

Answer 26

oh yeah, sorry same thing

Answer 27

I admit my error, but I still don't think that your solution is the intended method

Answer 28

yeah thats what i mean, you can individually add up those terms, i and i^2

Answer 29

gotchya, and I like your proof, but it is not by induction ;)

Answer 30

so 1(2) + 2(3) + 3(4) + n(n + 1) + (n+1)(n+2) will somehow yield the desired result?

Answer 31

meaning ((n+1)(n + 2)(n + 3))/3

Answer 32

\[\sum_{i=1}^{n}i(i+1)={n(n+1)(n+2)\over3}\]we have to show that this implies that\[\sum_{i=1}^{n+1}i(i+1)={(n+1)((n+1)+1)((n+1)+2)\over3}\](i.e. we get the same formula as we do with n, but replaced with n+1)

Answer 33

keep getting processing errors, i think I'm so close to understanding

Answer 34

damn, forgot about the error thing, so look you are very close
he have the sum up to n
1(2)+2(3)+...+n(n+1)=n(n+1)(n+2)/3
now the next term to both sides gives
1(2)+2(3)+...+n(n+1)+(n+1)(n+2)=n(n+1)(n+2)/3+(n+1)(n+2)
the left side is clearly true because all we did was add the next term

so we need to show through algebra that

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

and it will be proven

Answer 35

we have*

Answer 36

yessssss, i get it now, in essence, we don't need to worry about the LEFT side, but rather, add one to the right side

Answer 37

exactly!
glad you got the concept :)

Answer 38

keep up the good work

Answer 39

likewise!