[tex]\bf 4.\ \ \it 1^3=\dfrac{1^2\cdot2^2}{4} \Rightarrow 1=1\ (A)\\ \\ \\ Presupunem\ adev\breve arat\breve a:\\ \\ p(k):\ \ 1^3+2^3+3^3+\ ...\ +k^3=\dfrac{k^2(k+1)^2}{4}\\ \\ \\ Ar\breve at\breve am\ c\breve a\ este\ adev\breve arat\breve a:\\ \\ p(k+1):\ \ 1^3+2^3+3^3+\ ...\ +k^3+(k+1)^3=\dfrac{(k+1)^2(k+2)^2}{4}\\ \\ \\ 1^3+2^3+3^3+\ ...\ +k^3+(k+1)^3=\dfrac{k^2(k+1)^2}{4}+(k+1)^3\\ \\ \\ =\dfrac{k^2(k+1)^2+4(k+1)^3}{4}=\dfrac{(k+1)^2(k^2+4k+4)}{4}=\dfrac{(k+1)^2(k+2)^2}{4}[/tex]
Am demonstrat, prin inducție matematică, egalitatea dată.
