Question 1182883
{{{T[n] = C[2]^(n+1) = (n(n+1))/2}}}
{{{Q[n] = n^2}}}
{{{P[n] = (3n^2-n)/2}}}

(a)  {{{3P[n] = 3*((3n^2 - n)/2) = (1/2)*(3n)*(3n-1) = C[2]^(3n) = T[3n-1]}}}

(b)  {{{P[n]-Q[n] = (3n^2-n)/2 - n^2 = (n^2 - n)/2 = (n(n-1))/2 = 
C[2]^n = T[n-1]}}}

===>  {{{P[3n] - Q[3n] = T[3n-1]}}}
===>  {{{P[3n] - T[3n-1] = Q[3n]}}}  
Therefore,  {{{P[3n] - 3P[n] = Q[3n]}}} , from part (a).