THAM KHẢO NHA 

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CMR A = 1/5 + 1/6 + 1/7 + ... + 1/17 < 2

A = 1/5 + 1/6 + 1/7 + ... + 1/17

Vì 1/6 < 1/7 < 1/8 < 1/9 < 1/5 và 1/10 < 1/11 < 1/12 < 1/13 < 1/14 <1/15 < 1/16 < 1/17 < 1/8 nên:

A = (1/5 + 1/6 + 1/7 + 1/8 + 1/9) + (1/10 + 1/11 + 1/12 + 1/13 + 1/14+ 1/15 + 1/16 + 1/17)

A < (1/5 + 1/5 + 1/5 + 1/5 + 1/5) + (1/8 + 1/8 + 1/8 + 1/8 + 1/8+1/8 + 1/8 + 1/8)

A < 1 + 1

A < 2

Vậy: A < 2 (đpcm)

a>

\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000

ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )

1/100^2<1/2

=>A<1

oke

dpcm là gì vậy các bồ

dpcm là điều phải chứng minh nha

b,A= \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)

\(=(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{40})+(\dfrac{1}{41}+...+1...\)
\(=(\dfrac{20}{20.21}+\dfrac{21}{21.22}+...+\dfrac{39}{39.40})+(40/...\)
\(20(\dfrac{1}{20.21}+\dfrac{1}{21.22}+...\dfrac{1}{39.40})+40(\dfrac{1}{40}...\)
\(20(\dfrac{1}{20}-\dfrac{1}{40})+40(\dfrac{1}{40}-\dfrac{1}{60})>\dfrac{11}{15}\)
Lại có \(A<40(\dfrac{1}{20.21}+...\dfrac{1}{39.40})+60(\dfrac{1}{40.41}+...+...\)
\(=40(\dfrac{1}{20}-\dfrac{1}{40})+60(\dfrac{1}{40}-\dfrac{1}{60})<\dfrac{3}{2}\)

=> \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)

a,\( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\)

= \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+...+ \dfrac{1}{196} < \dfrac{1}{2^2-1}+ \dfrac{1}{4^2-1}+ \dfrac{1}{6^2-1}+...+ \dfrac{1}{14^2-1}\)

= \( \dfrac{1}{1.3}+ \dfrac{1}{3.5}+ \dfrac{1}{5.7}+...+ \dfrac{1}{13.15}\)

= \( \dfrac{1}{2}(1- \dfrac{1}{3}+ \dfrac{1}{3}- \dfrac{1}{5}+ \dfrac{1}{5}- \dfrac{1}{7}+ \dfrac{1}{7}-...- \dfrac{1}{13}+ \dfrac{1}{13}- \dfrac{1}{15})\)

= \( \dfrac{1}{2}(1- \dfrac{1}{15})< \dfrac{1}{2}\)

Vậy \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\) \(<\dfrac{1}{2} \)

hình như phân số cuối  phải là 1/324

nếu là 1/324 thì tớ giải nè:

A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324

= 1/4.(1+1/2^2+1/3^2+1/4^2+1/5^2+1/6^2+1/7^2+1/8^2+1/9^2)                                                    <1/4.(1+1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)

= 1/4.(1+1-1/9)

= 1/4.17/9 = 17/36<18/36 = 1/2

=> A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324<1/2

CHO ĐÚNG NHA!!!!!!!!!!!!!!!!

Đặt S=1/4+1/16+1/36+...+1/10000

        S= 1/4x(1+1/4+1/9+...+1/2500)

        S= 1/4x(1+1/2x2+1/3x3+...+1/50x50)

S< 1/4x(1+1/1x2+1/2x3+...1/49x50)

S< 1/4x(1+1-1/2+1/2-1/3+....+1/49-1/50)

S< 1/4x(1+1-1/50)

S< 1/4x(2-1/50)<2/4(2/4=1/2)

S< 1/2

S=\(\frac{1}{4}\)(1+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}+...+\frac{1}{50^2}\)

S<\(\frac{1}{4}\)(1+\(\frac{1}{2.1}\)+\(\frac{1}{3.2}+...+\frac{1}{50.49}\))

S<\(\frac{1}{4}\)(1+1−\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\))

S<\(\frac{1}{4}\)(2−\(\frac{1}{50}\))<\(\frac{2}{4}\)=\(\frac{1}{2}\)(đpcm)