Toán 1+2+2²+2³+………………………………………………………………………………….2mũ101 so sanh N=5.2mu100 19/09/2021 By Reese 1+2+2²+2³+………………………………………………………………………………….2mũ101 so sanh N=5.2mu100
Ta đặt $A = 1 + 2^2 + \cdots + 2^{101}$ Khi đó $2A = 2 + 2^2 + \cdots + 2^{102}$ Vậy $A = 2A – A = (2 + 2^2 + \cdots + 2^{102}) – (1 + 2^2 + \cdots + 2^{101})$ $= 2^{102} – 1$ Ta sẽ so sánh $2^{102} – 1$ và $5.2^{100}$ Ta xét $\dfrac{2^{102}-1}{5.2^{100}} = \dfrac{2^{102}}{5.2^{100}} – \dfrac{1}{5.2^{100}} = \dfrac{4}{5} – \dfrac{1}{5.2^{100}} < \dfrac{4}{5} < 1$ Vậy $2^{102}-1 < 5.2^{100}$ Trả lời
Ta đặt
$A = 1 + 2^2 + \cdots + 2^{101}$
Khi đó
$2A = 2 + 2^2 + \cdots + 2^{102}$
Vậy
$A = 2A – A = (2 + 2^2 + \cdots + 2^{102}) – (1 + 2^2 + \cdots + 2^{101})$
$= 2^{102} – 1$
Ta sẽ so sánh $2^{102} – 1$ và $5.2^{100}$
Ta xét
$\dfrac{2^{102}-1}{5.2^{100}} = \dfrac{2^{102}}{5.2^{100}} – \dfrac{1}{5.2^{100}} = \dfrac{4}{5} – \dfrac{1}{5.2^{100}} < \dfrac{4}{5} < 1$
Vậy $2^{102}-1 < 5.2^{100}$