Question

A=2⁰+2¹+2²+2³+2⁴+2⁵+2⁶...+2²⁰⁰³ se divide cu 7​

Answer 1

Explicație pas cu pas:

$\bf A=2^0 + 2^1 +2^2 +2^{3}+2^{4}+2^{5} + ....+2^{2003}$

$\bf A=\Big(2^0 + 2^1 +2^2\Big)+\Big(2^{3}+2^4+2^{5}\Big)+ ...+\Big(2^{2001}+2^{2002}+2^{2003}\Big)$

$A=\Big(1+ 2 +4\Big)+2^{3}\cdot\Big(2^{3-3}+2^{4-3}+2^{5-3}\Big)+ ...+2^{2001}\cdot\Big(2^{2001-2001}+2^{2002-2001}+2^{2003-2001}\Big)$

$\bf A=7+2^{3}\cdot\Big(2^{0}+2^1 +2^2\Big)+ ...+2^{2001}\cdot\Big(2^{0}+2^1 +2^2\Big)$

$\bf A=7+2^{3}\cdot\Big(1+ 2 +4\Big)+ ...+2^{2001}\cdot\Big(1+ 2 +4\Big)$

$\bf A=7+2^{3}\cdot 7+ ...+2^{2001}\cdot 7$

$\red{\boxed{~\bf A=7\cdot\Big(2^{0}+2^{3}+2^{6}+... +2^{2001}\Big)~~\vdots~~7~}}$

$==pav38==$