Lời giải:

\(A = \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{201 4^{2}}\)

\(< \frac{1}{4} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{2013.2014}\)

\(= \frac{1}{4} + \frac{3 - 2}{2.3} + \frac{4 - 3}{3.4} + . . . . + \frac{2014 - 2013}{2013.2014}\)

\(= \frac{1}{4} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . . + \frac{1}{2013} - \frac{1}{2014}\)

\(= \frac{1}{4} + \frac{1}{2} - \frac{1}{2014}\)

\(< \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\)

Ta có đpcm.

Lần sau bạn lưu ý gõ đề bằng công thức toán (biểu tượng \(\sum\) góc trái khung soạn thảo)

Cho A = 1/2^2 + 1/3^2 + 1/4^2 + ......... + 1/2014^2. Chứng tỏ A<3/4 

Lưu ý: Dấu ^ là dấu mũ nhaaa:3 Các cậu giải giúp tớ vớiiii:4

Ta có: \(\frac{1}{3^{2}} < \frac{1}{2.3}\)

           \(\frac{1}{4^{2}} < \frac{1}{3.4}\)

            .....................

            \(\frac{1}{201 4^{2}} < \frac{1}{2013.2014}\)

\(\Rightarrow A < \frac{1}{2^{2}} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{2013.2014}\)

Đặt \(B = \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{2013.2014}\)

           \(= \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . . + \frac{1}{2013} - \frac{1}{2014}\)

             \(= \frac{1}{2} - \frac{1}{2014} < \frac{1}{2}\)

\(\Rightarrow A < \frac{1}{2^{2}} + \frac{1}{2} = \frac{3}{4}\)

\(\text{Ta}\&\text{nbsp};\text{c} \overset{ˊ}{\text{o}} :\&\text{nbsp}; n^{2} > n^{2} - 1 = \left(\right. n - 1 \left.\right) \left(\right. n + 1 \left.\right)\)

\(\Rightarrow \frac{1}{n^{2}} < \frac{1}{\left(\right. n - 1 \left.\right) \left(\right. n + 1 \left.\right)} = \frac{1}{2} \left(\right. \frac{1}{n - 1} - \frac{1}{n + 1} \left.\right)\)

\(A = \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . . + \frac{1}{201 4^{2}} < \frac{1}{1.3} + \frac{1}{2.4} + \frac{1}{3.5} + . . . + \frac{1}{2013.2015}\)

\(= \frac{1}{2} \left(\right. 1 - \frac{1}{3} \left.\right) + \frac{1}{2} \left(\right. \frac{1}{2} - \frac{1}{4} \left.\right) + \frac{1}{2} \left(\right. \frac{1}{3} - \frac{1}{5} \left.\right) + . . . + \frac{1}{2} \left(\right. \frac{1}{2013} - \frac{1}{2015} \left.\right)\)

\(= \frac{1}{2} \left(\right. 1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + . . . + \frac{1}{2013} - \frac{1}{2015} \left.\right)\)

\(= \frac{1}{2} \left(\right. 1 + \frac{1}{2} - \frac{1}{2014} - \frac{1}{2015} \left.\right)\)

\(= \frac{1}{2} \left(\right. \frac{3}{2} - \frac{1}{2014} - \frac{1}{2015} \left.\right)\)

\(= \frac{3}{4} - \frac{1}{2} \left(\right. \frac{1}{2014} + \frac{1}{2015} \left.\right) < \frac{3}{4}\)

Vậy .............

A=1/100 mũ 2 +1/101 mũ 2 +...+1/2013mũ2 +1/2014 mũ 2     hãy chứng tỏ a < 1/99

\(A = \frac{1}{10 0^{2}} + \frac{1}{10 1^{2}} + . . . + \frac{1}{201 3^{2}} + \frac{1}{201 4^{2}}\)

\(A < \frac{1}{99.100} + \frac{1}{100.101} + . . + \frac{1}{2012.2013} + \frac{1}{2013.2014}\)

\(A < \frac{1}{99} - \frac{1}{100} + \frac{1}{100} - \frac{1}{101} + . . . + \frac{1}{2012} - \frac{1}{2013} + \frac{1}{2013} - \frac{1}{2014}\)

\(A < \frac{1}{99} - \frac{1}{2014} < \frac{1}{99}\)

Vậy A<1/99

Cho A =  1/5 + 1/5^2 + 1/5^3 +....+ 1/5^2014 . Chứng tỏ rằng A < 1/4 

\(A = \frac{1}{5} + \frac{1}{5^{2}} + . . . . . . . . + \frac{1}{5^{2014}}\)

\(\Rightarrow 5 A = 1 + \frac{1}{5} + . . . . . . . . . . . + \frac{1}{5^{2013}}\)

\(\Rightarrow 5 A - A = 1 + . . . . . . . . . . . + \frac{1}{5^{2013}} - \frac{1}{5} + . . . . . . . . . . . + \frac{1}{5^{2014}}\)

\(\Rightarrow 4 A = 1 - \frac{1}{5^{2014}}\)

\(\Rightarrow 4 A < 1 \Rightarrow A < \frac{1}{4}\)

=> 5A = 1 + 1/5 +...+1/5^2013

=>4A= 1- 1/5^2014

=> 4A< 1 => A < 1/4

CHo A=1/3^2+1/4^2+1/5^2+...+1/50^2. Chứng tỏ rằng 1/4<A<4/9

chứng tỏ rằng:(1/2 mũ 2+1/3 mũ 2 +1/3 mũ 2+1/4 mũ 2+1/5 mũ 2+1/6 mũ 2+1/7 mũ 2 +1/8 mũ 2+1/9 mũ 2+1/10 mũ 2)<1

\(A = \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{1 0^{2}}\)

\(< \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{9.10}\)

\(= \frac{2 - 1}{1.2} + \frac{3 - 2}{2.3} + \frac{4 - 3}{3.4} + . . . + \frac{10 - 9}{9.10}\)

\(= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . . + \frac{1}{9} - \frac{1}{10}\)

\(= 1 - \frac{1}{10} < 1\)

\(A = \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{1 0^{2}} A < \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + . . . + \frac{1}{9 \times 10} A < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . . + \frac{1}{9} - \frac{1}{10} = 1 - \frac{1}{10} A < \frac{9}{10} < 1 \Rightarrow A < 1\)

Cho A = 1/2^2+1/3^2+1/4^2+...+1/100^2. Hãy chứng  tỏ rằng 1/2<A <4/5

ko bit lm

ok

hok tot

Bài 1:So sánh 2014²⁰¹⁴ + 1/2014²⁰¹⁵ + 1 và 2014²⁰¹³ + 1/2014²⁰¹⁴ + 1. Bài 2: a) chứng tỏ rằng: D=1/2² + 1/3² + 1/4² +....+1/10² < 1. b)chứng tỏ rằng: E=1/101+1/102+...+1/299+1/300>2/3.C)chứng tỏ rằng: F=1/5+1/6+1/7+...+1/17 < 2

\(D = \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . . . . . + \frac{1}{1 0^{2}}\)

\(D < \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . . . . . + \frac{1}{9.10}\)

\(D < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . . . . + \frac{1}{9} - \frac{1}{10}\)

\(D < 1 - \frac{1}{10} \Leftrightarrow D < 1 \left(\right. đ p c m \left.\right)\)

1.

a, chứng tỏ

1/2^2+1/3^2+...+1/2017^2<1

b,1/4+1/16+1/36+1/64+1/100+1/144+...+1/10000<1/2

c,cho A=1/2^2+1/3^2...+1/9^2

chứng tỏ:2/5<a<8/9

d,chứng tỏ:A=1+1/2^2+...+1/100^2<1/3/4

e,chứng tỏ:1/2^2+1/3^2+...+1/100^2<1

a, Ta có: \(\frac{1}{2^{2}} < \frac{1}{1.2} ; \frac{1}{3^{2}} < \frac{1}{2.3} ; . . . ; \frac{1}{201 7^{2}} < \frac{1}{2016.2017}\)

\(\Rightarrow \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{201 7^{2}} > \frac{1}{1.2} + \frac{1}{2.3} + . . . + \frac{1}{2016.2017} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + . . . + \frac{1}{2016} - \frac{1}{2017} = 1 - \frac{1}{2017} < 1\)Vậy...

b, Đặt A = \(\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + . . . + \frac{1}{10000}\)

\(A = \frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + . . . + \frac{1}{10 0^{2}}\)

\(A = \frac{1}{2^{2}} \left(\right. 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right)\)

Đặt B = \(\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}}\)

Ta có: \(\frac{1}{2^{2}} < \frac{1}{1.2} ; \frac{1}{3^{2}} < \frac{1}{2.3} ; . . . . . ; \frac{1}{5 0^{2}} < \frac{1}{49.50}\)

\(\Rightarrow B < \frac{1}{1.2} + \frac{1}{2.3} + . . . + \frac{1}{49.50} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + . . . + \frac{1}{49} - \frac{1}{50} = 1 - \frac{1}{50} < 1\)

Thay B vào A ta được:

\(A < \frac{1}{4} \left(\right. 1 + 1 \left.\right) = \frac{1}{4} . 2 = \frac{1}{2}\)

Vậy....

c, Ta có: \(\frac{1}{2^{2}} > \frac{1}{2.3} ; \frac{1}{3^{2}} > \frac{1}{3.4} ; . . . . ; \frac{1}{9^{2}} > \frac{1}{9.10}\)

\(\Rightarrow A > \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{9.10} = \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . . + \frac{1}{9} - \frac{1}{10} = \frac{1}{2} - \frac{1}{10} = \frac{2}{5}\)(1)

Lại có: \(\frac{1}{2^{2}} < \frac{1}{1.2} ; \frac{1}{3^{2}} < \frac{1}{2.3} ; . . . . ; \frac{1}{9^{2}} < \frac{1}{8.9}\)

\(\Rightarrow A < \frac{1}{1.2} + \frac{1}{2.3} + . . . + \frac{1}{8.9} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + . . . + \frac{1}{8} - \frac{1}{9} = 1 - \frac{1}{9} = \frac{8}{9}\)(2)

Từ (1) và (2) suy ra \(\frac{2}{5} < A < \frac{8}{9}\)(đpcm)

d, chắc là đề sai

e, giống câu a

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\(\frac14+\frac19+\frac{1}{16}+\cdots+\frac{1}{2014^2^{}}<\frac14+\frac{1}{2.3}+\frac{1}{3.4}+\cdots+\frac{1}{2013.2014}=\frac14+\frac12-\frac13+\frac13-\frac14+\ldots+\frac{1}{2013}-\frac{1}{2014}=\frac14+\frac12-\frac{1}{2014}=\frac34-\frac{1}{2014}<\frac34\)

>

triều đại vua hùng

Nhà em có một con thú rật thông minh.Nó đã 12 tuổi.Nó là loài động vật bậc cao ngang một con người.Đó là em của em

ko biieets

Ta có tổng các chữ số của số hạng cuarM là:1+2+3+...+1000

*Ta có tổng quát 3 số tự nhiên liên tiếp:n+(n+1)+(n+2)=3n+3 chia hết cho 3 (n\(\in\) N)

Do đó (1+2+3)+(4+5+6)+....+(997+998+999) chia hết cho 3

Suy ra M chia 3 dư 1

Ta có:

M=1/2+3/2+(3/2)^2+....+(3/2)^2023

3/2M=3/4+(3/2)^2+(3/2)^3+....+(3/2)^2024

3/2M-M=3/2+(3/2)^2+(3/2)^3+....+(3/2)^2024-[1/2+(3/2)+(3/2)^2+...+(3/2)^2023)]

1/2M=(3/2)^2024-1/2

M=3^2024/2^2023-1

M=3^2024-2^2023/2^2023

Suy ra M-N=3^2024-2^2023/2^2023-3^2024/2^2023

M-N=-2^2023/2^2023

M-N=-1

Vậy M-N=-1