n ∑ i (i+3) = 1/3 n (n+1) (n+5) i=1 zixxx: −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 1.dla n=1 l=4=p prawda 2 dla nEN zaloz ind..... n+1 teza ind ∑ i(i+3) = 1/3(n+1) (n+2) (n+6) i=1 rozw: ∑ i(i+3) = ∑ i(i+3) + (n+1)( n+5) = 1/3 n (n+1) (n+5) + (n+1)(n+5)= 1/3(n+1)(n2 +5n ) + (n+1)(n+4) nie wychodzi..bardzo proszę o pomoc