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a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
\(\Rightarrowđpcm\)
d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
A = \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
A = \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
A = B - 2C ( ĐPCM )
Vậy A = B - 2C
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
a) S = 1.2 + 2.3 + 3.4 + ... + 99.100
S có thể được viết lại thành:
S = 1(2 - 0) + 2(3 - 1) + 3(4 - 2) + ... + 99(100 - 98)
= 1.2 - 0 + 2.3 - 1 + 3.4 - 2 + ... + 99.100 - 98
= (1.2 + 2.3 + 3.4 + ... + 99.100) - (0 + 1 + 2 + ... + 98)
Để tính tổng 1.2 + 2.3 + 3.4 + ... + 99.100, ta sử dụng công thức:
S = n(n+1)(2n+1)/6
Với n = 99, ta có:
S = 99.100.199/6 = 331650
Tính tổng 0 + 1 + 2 + ... + 98, ta sử dụng công thức:
S = n(n+1)/2
Với n = 98, ta có:
S = 98.99/2 = 4851
Do đó, S = 331650 - 4851 = 326799
b) B = 4924.12517.28−530.749.45529.162.748
B có thể được viết lại thành:
B = (4924.12517.28) / (530.749.45529.162.748)
B = (4924 / 530) . (12517 / 749) . (28 / 45529) . (162 / 162) . (748 / 748)
B = 9.17.28/45529 = 2^2 . 3^2 . 17 / 45529
B = 108 / 45529
c) C = (13+132+133+134).35+(135+136+137+138).39+...+(1397+1398+1399+13100).3101
C = (13(1 + 13 + 13^2 + 13^3)) . 3^5 + (13^5(1 + 13 + 13^2 + 13^3)) . 3^9 + ... + (13^97(1 + 13 + 13^2 + 13^3)) . 3^101
C = (1 + 13 + 13^2 + 13^3) . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)
C = 80 . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)
C = 80 . (13^5 . 3^4 . 3 + 13^9 . 3^8 . 3 + ... + 13^97 . 3^96 . 3)
C = 80 . (13^6 . 3^5 + 13^10 . 3^9 + ... + 13^98 . 3^97)
C = 80 . 3^5 (13^6 + 13^10 + ... + 13^98)
d) D = 3 - 3^2 + 3^3 - 3^4 + ... + 3^2017 - 3^2018
D = (3 - 3^2) + (3^3 - 3^4) + ... + (3^
- Ta có B-2C = (1/1+1/2+1/3+1/4+...+1/49+1/50) - (1/1+1/2+1/3+....+1/25)
= 1/26+1/27+1/28+...+1/50
- Ta có A= 1-1/1.2+1/3-1/4+1/5-1/6+.....+1/49-1/50
= (1+1/2+1/3+14+...+1/49)-(1/2+1/4+1/6+...+1/50)
=(1+1/2+1/3+...+1/50)-2(1/2+1/4+1/6+...+1/50)
= ( 1+1/2+...+1/50)-(1+1/2+1/3+...+1/25)
= 1/26+1/27+....+1/50
=> A=B-2C <ĐPCM>
Vậy A=B-2C
- mik giải hơi dài theo cách cổ điển ^_^ ^_^
1
Ta có :A=1/1.2+1/3.4+...+1/99.100=1/2+1/12+...+1/9900
7/12=1/2+1/12
Vì 1/2+1/12<1/2+1/12+...+1/9900
Nên: 7/12<A (1)
Lại có:A=1/1.2+1/3.4+...+1/99.100
=1-1/2+1/3-1/4+...+1/99-1/100
=(1-1/2+1/3)+(-1/4+1/5-1/6)+...+(-1/98+1/99-1/100)
5/6=1-1/2+1/3
vì: 1-1/2+1/3 < (1-1/2+1/3)+(-1/4+1/5-1/6)+...+(-1/98+1/99-1/100)
nên 5/6 < A (2)
Từ (1) và (2) suy ra 7/12<A<5/6
a)A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) > 1 / (1*2) + 1 / (3*4) = 1 / 2 + 1 / 12 = 7 / 12 ♦
A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100) =
(1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 <
1 - 1 / 2 + 1 / 3 = 5 / 6 ♥
♦, ♥ => 7 / 12 < A < 5 / 6
b)ta có:
1/1.2+1/3.4+1/5.6+...+1/49.50
=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50
=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)
=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2
=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)
=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50
hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50