Question

) Arătaţi că: a) A = 1/2007 + 2/2 + 3/3 se divide cu 16;
b) B = 7n + 7n+1 + 71+2 + 7n+3 se divide cu 10.

Answer 1

Răspuns:

$\color{CC0000}\huge \boxed{\bf A = 32 \: \vdots \: 16}$

$\color {DarkBlue}\huge\boxed{\bf B = {7}^{n} \cdot {2}^{2} \cdot \: {10}^{2} \: \: \vdots \: \: 10 }$

Explicație pas cu pas:

Bună!

$\bf (a) \: \: A = {1}^{2007} + {2}^{2} + {3}^{3}$

$\bf A = 1 + 4 + 27$

$\color {CC0000} \boxed{ \boxed{\bf A = 32 \: \vdots \: 16}}$

$\bf (b) \: \: B = {7}^{n} + {7}^{n + 1} + {7}^{n + 2} + {7}^{n + 3}$

$\bf B = {7}^{n} \cdot({7}^{n - n} + {7}^{n + 1 - n} + {7}^{n + 2 - n} + {7}^{n + 3 - n} )$

$\bf B = {7}^{n} \cdot({7}^{0} + {7}^{1} + {7}^{2} + {7}^{3} )$

$\bf B = {7}^{n} \cdot(1+ 7 + 49 + 343 )$

$\bf B = {7}^{n} \cdot 400$

$\bf B = {7}^{n} \cdot 4 \cdot \: 100$

$\color {DarkBlue} \boxed {\boxed{\bf B = {7}^{n} \cdot {2}^{2} \cdot \: {10}^{2} \: \: \vdots \: \: 10 }}$

==pav38==