Help with Induction??? (1 Viewer)

Heresy

Active Member
I'm not sure why but I'm having trouble with factorising it and proving it (n =k+1 - after you sub in n=k) - smh...

jathu123

Active Member
$\bg_white 1^2 +2^2 +3^2+ ... +k^2 =\frac{1}{6}k(k+1)(2k+1) \qquad (*)\\ \\ Prove true for n = k+1, ie: \\ 1^2 +2^2 +3^2+ ... +k^2 + (k+1)^2 =\frac{1}{6}(k+1)(k+2)(2k+3) \\ LHS = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2 \qquad from (*) \\ Now factorse \frac{1}{6}(k+1): \\ =\frac{1}{6}(k+1)\left (k(2k+1)+6(k+1) \right ) \\ \\ = \frac{1}{6}(k+1)(2k^2+7k+6) \\ \\ =\frac{1}{6}(k+1)(k+2)(2k+3) = RHS \\ \\ \therefore true for n=k+1$

fan96

617 pages
We have

$\bg_white \frac 16 k(k+1)(2k+1) = \frac 1 6(2k^3+3k^2+k)$

and we want to obtain

$\bg_white \frac 16 (k+1)(k+2)(2k+3) = \frac 1 6(2k^3+9k^2+13k+6)$

Now,

$\bg_white \left[\frac 16 k(k+1)(2k+1) \right]+ (k+1)^2 = \frac 1 6(2k^3+3k^2+k) + \frac 1 6(6k^2+12k+6)$

...

Heresy

Active Member
Thank you both - I realised what I did wrong just before I came back on the thread - I'm so stupid. Sorry to you both...