\(A=1\cdot2+2\cdot3+...+n\left(n+1\right)\\ \Rightarrow3\cdot A=1\cdot2\cdot3+2\cdot3\cdot3+...+n\left(n+1\right)\cdot3\)

\(\Rightarrow3\cdot A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+n\left(n+1\right)\left(n+2-\left(n-1\right)\right)\)\(\Rightarrow3\cdot A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)\(\Rightarrow3\cdot A=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow A=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)

Ta có: \(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)

              \(=\left(2+2^2\right)+2^2\cdot\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

              \(=\left(2+2^2\right)\cdot\left(1+2^2+...+2^{98}\right)\)

              \(=6\cdot\left(1+2^2+...+2^{98}\right)\)

\(\Rightarrow\) A chia hết cho 6

Vì 6 mũ số nào thì luôn tận cùng là 6 \(\left(6^1=6,6^2=36,6^3=216,...\right)\)

nên chữ số tận cùng của \(6^{200}\)  là 6

a, \(R_1ntR_2\)\(\Rightarrow R_{12}=R_1+R_2=6+6=12\left(\Omega\right)\)

\(R_{12}//R_3\Rightarrow R_{tđ}=\dfrac{R_{12}\cdot R_3}{R_{12}+R_3}=\dfrac{12\cdot9}{12+9}=\dfrac{36}{7}\left(\Omega\right)\)

b, \(U_{AB}=I\cdot R_{tđ}=1,5\cdot\dfrac{36}{7}=\dfrac{54}{7}\left(V\right)\)

\(R_{12}//R_3\Rightarrow U_{12}=U_3=U_{AB}=\dfrac{54}{7}\left(V\right)\)

\(\Rightarrow I_{12}=\dfrac{U_{12}}{R_{12}}=\dfrac{\dfrac{54}{7}}{12}=\dfrac{9}{14}\left(A\right)\)

Mà \(R_1ntR_2\Rightarrow I_1=I_2=I_{12}=\dfrac{9}{14}\left(A\right)\)

\(\Rightarrow U_1=I_1\cdot R_1=\dfrac{9}{14}\cdot6=\dfrac{27}{7}\left(V\right)\\ U2=I_2\cdot R_2=\dfrac{27}{7}\left(V\right)\)

Vậy \(U_1=U_2=\dfrac{27}{7}\left(V\right);U_3=\dfrac{54}{7}\left(V\right)\)

Áp dụng bđt côsi với các số không âm ta có:

\(a+b\ge2\sqrt{ab}\\ a+1\ge2\sqrt{a}\\ b+1\ge2\sqrt{b}\)

=> \(a+b+a+1+b+1\ge2\sqrt{ab}+2\sqrt{a}+2\sqrt{b}\)

=> \(2a+2b+2\ge2\sqrt{ab}+2\sqrt{a}+2\sqrt{b}\)

=> \(a+b+1\ge\sqrt{a}+\sqrt{b}+\sqrt{ab}\) ( ĐPCM)