Question

A=1+3+3 la puterea 2+3 la puterea 3+3 la puterea 4+......+3 la puterea 2003 divizibil cu 4​

Answer 1

Ai demonstratia mai jos

Explicație pas cu pas:

Salutare!

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

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

$\bf A =(1+3)+{3}^{2}\cdot({3}^{0}+{3}^{1})+{3}^{4}\cdot({3}^{0}+{3}^{1})+ ...... +{3}^{2002}\cdot({3}^{0}+{3}^{1})$

$\bf A =4+{3}^{2}\cdot(1+3)+{3}^{4}\cdot(1+3)+ ...... +{3}^{2002}\cdot(1+3)$

$\bf A =4+{3}^{2}\cdot 4+{3}^{4}\cdot 4+ ...... +{3}^{2002}\cdot 4$

$\bf A =4 \cdot(1+{3}^{2}+{3}^{4}+ ...... +{3}^{2002}) \: \vdots \: 4$

==pav38==