Question 7468
i think some of your working is wrong.


Anyway, what i would do is the following:


assuming that {{{1^2 + 2^2 + ...}}} + {{{k^2 = (k(k+1)(2k+1))/6}}} then we have to prove that {{{1^2 + 2^2 + ...}}} + {{{k^2 + (k+1)^2}}}, which is {{{(k(k+1)(2k+1))/6 + (k+1)^2}}} will be equivalent to {{{((k+1)((k+1)+1)(2(k+1)+1))/6}}}


Is this OK?


Right... {{{(k(k+1)(2k+1))/6 + (k+1)^2}}} becomes

(2k^3+9k^2+13k+6)/6


Now I am looking for one of the factors to be (k+1), so divide this into the numerator...it divides exactly, as I would expect, to leave {{{2k^2+7k+6}}} which then factorises to give, fully,


{{{((k+1)(k+2)(2k+3))/6}}}


Right, nearly there.

The first term is fine.
The second term can be written as (k+1) + 1
the third term can be written as 2(k+1)+1... just make sure you are happy that this is equivalent to 2k+3.


so, we have proved that {{{1^2 + 2^2 + ...}}} + {{{k^2 + (k+1)^2}}} gives a formula of the same form as that for {{{1^2 + 2^2 + ...}}} + {{{k^2}}} by adding on the next term, namely (k+1), therefore QED.


jon.