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A = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{308}\)+ \(\frac{1}{309}\)
B = \(\frac{308}{1}\)+ \(\frac{307}{2}\)+ \(\frac{306}{3}\)+\(\frac{3}{306}\) + \(\frac{2}{307}\)+ \(\frac{1}{308}\)
=> B = \(\frac{309-1}{1}\)+ \(\frac{309-3}{3}\)+... + ( 309 ... )
=> B = 309 + 309 . ( \(\frac{1}{2}\) + \(\frac{1}{3}\)+... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\)- \(\frac{1}{1}\)+ \(\frac{2}{2}\)+ ... + \(\frac{308}{308}\)+ \(\frac{309}{309}\)
=> B = 309 . ( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ ... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\))
=> \(\frac{A}{B}\)= \(\frac{1}{309}\)
Ta có :
\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+...+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)
\(B=\left(\frac{307}{2}+1\right)+\left(\frac{306}{3}+1\right)+...+\left(\frac{3}{306}+1\right)+\left(\frac{2}{307}+1\right)+\left(\frac{1}{308}+1\right)+1\)
\(B=\frac{309}{2}+\frac{309}{3}+...+\frac{309}{306}+\frac{309}{307}+\frac{309}{308}+\frac{309}{309}\)
\(B=309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{306}+\frac{1}{307}+\frac{1}{308}+\frac{1}{309}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}}{309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{308}+\frac{1}{309}\right)}\)
\(\frac{A}{B}=\frac{1}{309}\)
ĐặtA= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\)
\(\frac{1}{A}=1\div\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\right)\)
=> \(\frac{1}{A}=2+3+4+...+308+309\)
=>Ta có: Số các số hạng là:(309-2)/1+1=308(số hạng)
Tổng của\(\frac{1}{A}\)là:\(\frac{\left(309+2\right).308}{2}\)=47894
=> \(\frac{1}{A}=47894\)
=>\(A=\frac{1}{47894}\)
Chúc bạn học tốt!
\(B=308/1+307/2+306/3+...+1/308 \)
\(B=308+307/2+306/3+...+1/308\) chia số 308 thành 308 số 1
B=307/2+1+306/3+1+...+1/308+1+1
B=309/2+309/3+309/4+...+309/308+309/309
B=309(1/2+1/3+1/4+...+1/309)=309A
Suy ra A/B=1/309
=(1/2+1/31/4...1/307/1/3081/309)/(309-1/1+309-2/2+...+309-307/307+309-308/308)
=(1/21/31/4...1/3071/3081/309)/(309/1-1+309/2-1+...+309/307-1+309/308-1)
=(........................................)/(309/309309/2309/3...309/307+309/308)
=(........................................)/[309x(1/309+1/308+...+1/41/31/2)]
Thấy tử và mẫu giống nhau thì ta rút:
=1/309
a. Ta có :
B = 308/1 + 307/2 +306/3+....+1/308
B = (1+1+....+1) + 307/2 + ....+ 1/308
B = (1 + 307/2) + (1+306/3) + ...+ (1+ 1/308) + 1
B = 309/2 + 309/3 + ....+ 309/308 + 309/309
B = 309.(1/2 + 1/3 + ....+1/309)
Vậy A/B: 1/2 + 1/3 + ... + 1/309 / 308/1 + 307/2 +....+ 2/307+1/308
A/B = 1/2 + 1/3 +... + 1/309 / 309.(1/2 + 1/3 + ....+1/309)
A/B = 1/309
b.7/10.11 + 7/11.12 + .... +7 /69.70
= 7. (1/10.11+1/11.12 + ...+ 1/69.70)
= 7.(1/10-1/11+1/11-1/12+....+1/69-1/70)
= 7.(1/10 - 1/70)
= 7. 3/35
= 3/5
Bài 2:
a, S = 1/11 + 1/12 + .. +1/20 với 1/2
SỐ số hạng tổng S: [20 - 11]: 1 + 1 = 10 số
mà 1/11 > 1/20
1/12 > 1/20
.........................
1/20 = 1/20
=> 1/11 + 1/12 + ... + 1/20 > 1/20 . 10 => S > 1/2
b, B = 2015/2016 + 2016/2017 và C = 2015+2016/2016+2017
Dễ dàng ta thấy: C = 4031/4033 < 1
B = 2015/2016 + 2016/2017
B = 2015/2016 + [1/2016 + 4062239/4066272]
B = [2015/2016 + 1/2016] + 4062239/4066272]
B = 1 +4062239/4066272
=> B > 1
Vậy B > C
c, [-1/5]^9 và [-1/25]^5
ta có: 255 = [52]5 = 52.5 = 510 > 59
=> [1/5]9 > [1/25]5
=> [-1/5]9 < [-1/25]5
d, 1/32+1/42+1/52+1/62 và 1/2
ta có: 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 = 1/9 + 1/16 + 1/25 + 1/36
mà: 1/9 < 1/8
1/16 < 1/8
1/25 < 1/8
1/36 < 1/8
=> 1/9+1/16+1/25+1/36 < 1/2
Vậy 1/32+1/42+1/52+1/62 < 1/2
Bài 1:
A = 3/4 . 8/9 . 15/16....2499/2500
A = [1.3/22][2.4/32]....[49.51/502]
A = [1.2.3.4.5...51 / 2.3.4....50][3.4.5...51 / 2.3.4...50]
A = 1/50 . 51/2
A = 51/100
B = 22/1.3 + 32/2.4 + ... + 502/49.51
B = 4/3.9/8....2500/2499
Nhận thấy B ngược A => B = 100/51 [cách tính tương tự tính A]
Bài 2:
a. S = 1/11+1/12+...+1/20 và 1/2
Số số hạng tổng S: [20 - 11]: 1 + 1 = 10 [ps]
ta có: 1/11 > 1/20
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)