Ta có :
\(P=1+2+2^2+.........................+2^{14}\)
\(\Rightarrow P=\left(1+2+2^2+2^3+2^4\right)+........+\left(2^9+...+2^{14}\right)\)
\(\Rightarrow P=2\left(1+2+....+2^4\right)+.....+2^{10}\left(1+2+...+2^4\right)\)
\(\Rightarrow P=2.31+......+2^{10}.31\)
\(\Rightarrow P=31\left(2+...+2^{10}\right)⋮31\)
\(\rightarrowđpcm\)
Ta có:
P=1+2+22+23+...+213+214
=(1+2+22+23+24)+(25+26+27+28+29)+(210+211+212+213+214)
=31+25(1+2+22+23+24)+210(1+2+22+23+24)=31+25.31+210.31\(⋮\)31
Hình như bài này phải giải là:
P = 1 + 2 + 22 + 23 + ... + 213 + 214
=> P = ( 1 + 2 + 22 + 23 + 24 ) + ( 25 + 26 + 27 + 28 + 29 ) + ( 210 + 211 + 212 + 213 + 214 )
= 31 + 25. ( 1 + 2 + 22 + 23 + 24 ) + 210. ( 1 + 2 + 22 + 23 + 24 )
= 31 + 25 . 31 + 210 . 31
= 31 . ( 1 + 25 + 210 ) chia hết cho 3.