E.694. Să se demonstreze că A=4⋅15n+4⋅134n+342n+1 ⋮ 7, ∀n∈N.A=4\cdot 15^n+4 \cdot 134^n + 34^{2n+1} ~\vdots~7, ~\forall n\in \N.A=4⋅15n+4⋅134n+342n+1 ⋮ 7, ∀n∈N.
A=4⋅(14+1)n+4⋅(133+1)n+(35−1)2n+1=A=4 \cdot(14+1)^n+4 \cdot (133+1)^n + (35-1)^{2n+1}=A=4⋅(14+1)n+4⋅(133+1)n+(35−1)2n+1= =4(M14⏟M7+1)+4(M133⏟M7+1)+M35⏟M7−1==4( \underbrace{M_{14}}_{M_7}+1) + 4 (\underbrace{M_{133}}_{M_7}+1) + \underbrace{M_{35}}_{M_7}-1==4(M7M14+1)+4(M7M133+1)+M7M35−1= =4M4+4+4M7+4+M7−1=M7+7=M7.=4M_4+4+4M_7+4+M_7-1=M_7+7 = M_7.=4M4+4+4M7+4+M7−1=M7+7=M7.