Question

sa se demonstreze ca 2(1+3+3²+.....+3⁸) < 3⁹​

Answer 1

$2(1 + 3 + 3 {}^{2} + ... + 3 {}^{8} ) < 3 {}^{9}$

Avem o progresie geometrică cu rația 3

Suma unei progresii geometrice are formula:

$\frac{b1(q {}^{n} - 1)}{q - 1}$

$2 \times \frac{1 \times 3 {}^{9} - 1}{3 - 1} < 3 {}^{9}$

$2 \times \frac{3 {}^{9} - 1}{2} < 3 {}^{9}$

2 cu 2 se simplifica

$3 {}^{9} - 1 < 3 {}^{9}$

Reducem 3⁹ si obținem:

$- 1 < 0$

relatie adevarata

Answer 2

3⁹ = 3⁸•3 = 3⁸•(2+1) =

= 2•3⁸+3^8 = 2•3⁸+3⁷•3 =

= 2•3⁸+3⁷•(2+1) = ... =

= 2•(3⁸+3⁷+3⁶+...+3²)+3•(2+1) =

= 2•(3⁸+3⁷+3⁶+...+3²)+3•2+3 =

= 2•(3⁸+3⁷+3⁶+...+3²+3) + 3 =

= 2•(3⁸+3⁷+3⁶+...+3²+3+1)+1 > 2•(1+3+3²+.....+3⁸)

⇔ 1 > 0 (A)

⇒ 2•(1+3+3²+.....+3⁸) < 3⁹​