Help with Induction??? (1 Viewer)

Heresy · Dec 10, 2018

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 · Dec 10, 2018

fan96 · Dec 10, 2018

We have

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

and we want to obtain

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

Now,

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

...

Heresy · Dec 10, 2018

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...

Help with Induction??? (1 Viewer)

Heresy

Active Member

jathu123

Active Member

fan96

617 pages

Heresy

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)