Chứng tỏ rằng:B = 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + 1/2^6 + 1/2^7 + 1/2^8 < 1 18/10/2021 Bởi Melody Chứng tỏ rằng:B = 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5 + 1/2^6 + 1/2^7 + 1/2^8 < 1
Ta có $B = \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^8}$ $< \dfrac{1}{1\times 2} + \dfrac{1}{2\times 3} + \cdots + \dfrac{1}{7\times 8}$ $= 1 – \dfrac{1}{2} + \dfrac{1}{2} – \dfrac{1}{3} + \cdots + \dfrac{1}{7} – \dfrac{1}{8}$ $= 1 – \dfrac{1}{8} < 1$ Vậy $B < 1$. Bình luận
1+1=2
Ta có
$B = \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{2^8}$
$< \dfrac{1}{1\times 2} + \dfrac{1}{2\times 3} + \cdots + \dfrac{1}{7\times 8}$
$= 1 – \dfrac{1}{2} + \dfrac{1}{2} – \dfrac{1}{3} + \cdots + \dfrac{1}{7} – \dfrac{1}{8}$
$= 1 – \dfrac{1}{8} < 1$
Vậy $B < 1$.