Question

Qual é a soma da PG finita (1,2...512)? Preciso pra agr

Answer 1

PG (1,2, ... , 512)

a1 = 1, a2 = 2, q = 2/1 = 2 , an = 512

$a_{n} = a_{1}. q^{n-1}$

$512 = 1. 2^{n-1} \\\\ \frac{512}{1} = 2^{n}. 2^{-1} \\\\ 512 = 2^{n}. \frac{1}{2} \\\\ \frac{512}{ \frac{1}{2} }= 2^{n} \\\\ 1024 = 2^{n} \\\\ 2^{10} = 2^{n} \\\\ 10 = n ---\ \textgreater \ n = 10$

Cálculo da soma:

$S_{n}= \frac{ a_{1}( q^{n}-1 ) }{q-1} \\\\ S_{10}= \frac{1.( 2^{10}-1 )}{2-1} \\\\ S_{10}= \frac{1024-1}{1} \\\\ S_{10}= 1023$

Portanto: a soma da PG dada é 1023.