Question

Prove that the sum of x squares is \[\frac{3x^3 + 2x^2 + x}6\]

Answer 1

is it possible to prove this using some method other than induction?

Answer 2

yes there is

Answer 3

what would it be?

Answer 4

Ha ha ha ha..

Hey lgba how are you???

Answer 5

How you posted two questions lgba ??

Answer 6

He has magical powers. lol

Answer 7

Yeah I know..

Answer 8

Hey parth how are you??

Answer 9

Hi gsRttN, I'm great and you?

Answer 10

Yeah I am also fine..

How is your summer school going ??

Answer 11

Summer school?

Answer 12

Yeah your school name...

Answer 13

http://answers.yahoo.com/question/index?qid=20080613040132AAhpkMi

Answer 14

The solution is very large..

Answer 15

It should be \[\frac{2x^3 + 3x^2 + x}6\] not\[\frac{3x^3 + 2x^2 + x}6\]@ihbasallote

Answer 16

How do you know, gsRttN?

Answer 17

@gsRttN

Answer 18

Can't I know this??? @ParthKohli

Answer 19

The solution is very large to be written here, so I have given the link...

Answer 20

lol hi waterineyes.

Answer 21

@gsRttN: why did you make a new account?

Answer 22

If you don't like induction (I don't either, sometimes there is no choice) you can make use of the following:

Sum to n [(i + 1)k - i^k] = (n+1)^k -1

Setting k to 2, 3 etc will produce formulas (with some algebra, of course) for sums of first, second, third powers....

Answer 23

...isn't that induction too?

Answer 24

U know I'm fond of explicit constructions, right?

Answer 25

So to do with k = 2 ->

(i+1)^2 = 2^2 + 3^2 +.... +(n-1)^2 + n^2 + (n+1)^2
i^2=1^2+ 2^2 + ........ + n^2 and subtract ->

-1 + (n+1)^2 which is twice the usual formula for sum of integers 1 to n.

Answer 26

However, induction is easier, just less informative....

Answer 27

you can prove one from the previous one

Answer 28

that is to say, if you know
\[\sum_{k=1}^nk\] then you know
\[\sum_{k=1}^nk^2\] by algebra
and also if you know
\[\sum_{k=1}^nk^2\] then algebra gives you
\[\sum_{k=1}^nk^3\] etc
would be happy to show you when you are here

Answer 29

btw if you know the answer is a polynomial you don't even need induction, just compute for three values and you are done

Answer 30

i'm confused...isn't proving using summation considered induction?

Answer 31

not at all, and it usually is totally unnecessary
inductions proofs for summation formulas are trivial, since reducing to the previous case is nothing more than saying "sum up to the second to last term"

Answer 32

is it possible to prove this without using those summations at all? honestly, i'm very bad with summation

Answer 33

but in any case let me outline how you can find
\[\sum_{k=1}^nk^2\] if you know
\[\sum_{k=1}^nk=\frac{n(n+1)}{n}\]

Answer 34

Sure, you can do it like I said but it is a lot of writing........

Answer 35

oh not at all

Answer 36

With summation symbols, it is much shorter

Answer 37

But usually gives younger students a headache:-)

Answer 38

hmm...maybe i should try learning summation first....

Answer 39

ok i have a recommendation for you then
work through this

Answer 40

the idea is to work from the previous case, so for example if you know
\[\sum_{k=1}^nk=\frac{n(n+1)}{2}\] then you consider
\[n^3=\sum_{k=1}^nk^3-(k-1)^3\]
do some algebra, and see that you get an expression in terms only of \(k^2\) and \(k\) then do a but more algebra to solve for the sum you want

Answer 41

and yes, you should learn summation first