/sci/ - Science & Math

>>16101889
i know nobody replied and probably nobody cares about this baby shit but my solution was assuming the proposition was true for [math]n=k-1[/math] as well, then doing
[math]f_{c+k+1} = f_{c+k} + f_{c+k-1}[/math]
[math]= (f_cf_{k+1} + f_{c-1}f_k) + ((f_cf_{k} + f_{c-1}f_{k-1}))[/math], using our fancy new assumption
[math]= f_c(f_{k+1} + f_k) + f_{c-1}(f_k + f_{k-1})[/math]
[math]= f_c(f_{k+2}) + f_{c-1}(f_{k+1})[/math],
so if it's true for [math]n = k, n = k-1[/math] then it's true for [math]n = k+1[/math].

when [math]n=2[/math] the propo says we should get [math]f_{c+2} = f_cf_3 + f_{c-1}f_2 = 2f_c + f_{c-1}[/math], which we arrive at via [math]f_{c+2} = f_{c+1} + f_c = (f_c + f_{c-1} + f_c = 2f_c + f_{c-1}[/math], so it's true for [math]n = 2[/math].

it's true for [math]n=2[/math] and [math]n=2-1=1[/math], so it must be true for [math]n=3[/math], and therefore [math]n=5, 6, ...[/math] for all positive [math]\mathbb{Z}[/math] Q-E-mothafuckin-D.

i didn't intuit that you could do induction using two assumptions if you satisfied the larger base case until just now!
[math][/math]