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19 tháng 8 2023

Each term of S is n!(n2 + n + 1) = n![n(n + 1) + 1] = n(n + 1)n! + n!

By definition, n(n + 1)n! + n! = n! + n(n + 1)!

Therefore, S can be simplified as

1! + 1.2! + 2! + 2.3! + ... + 100! + 100.101!

So \(\dfrac{S+1}{101!}=\dfrac{1+1!+1\cdot2!+2!+2\cdot3!+...+100!+100\cdot101!}{101!}\)

\(=\dfrac{2!+1\cdot2!+2!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)

\(=\dfrac{3!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)

\(=\dfrac{4!+3\cdot4!+4!+...+100!+100\cdot101!}{101!}\)

\(=...\)

\(=\dfrac{100!+99\cdot100!+100!+100\cdot101!}{101!}\)

\(=\dfrac{101!+100\cdot101!}{101!}\)

\(=1+100=101\)

Hence, \(\dfrac{S+1}{101!}=101\)

30 tháng 10 2019

a)  18 2 < 10 3

b)  3 2 + 4 2 < ( 3 + 4 ) 2

c)  100 2 + 30 2 < ( 100 + 30 ) 2

d)  a 2 + b 2 > ( a - b ) 2 với a ∈   N * ;   b ∈   N * .  

22 tháng 12 2023

a) \(3.5^2+15.2^2-26\div2\)

= 3.25 + 15.4 - 13

= 75 + 60 - 13

= 135 - 13

= 122

b) \(5^3.2-100\div4+2^3.5\)

= 125.2 - 25 + 8.5

= 250 - 25 + 40

= 225 + 40

= 265

c)\(6^2\div9+50.2-3^3.33\)

= 36 : 9 + 100 - 9.33

= 4 + 100 - 297

= 104 - 297

= -193

d)\(3^2.5+2^3.10-81\div3\)

= 9.5 + 8.10 - 27

= 45 + 80 - 27

= 125 - 27

= 98

e) \(5^{13}\div5^{10}-25.2^2\)

= 53 - 25.4

= 125 - 100

= 25

f) \(20\div2^2+5^9\div5^8\)

= 20 : 4 + 5

= 5 + 5

= 10

22 tháng 6 2023

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

25 tháng 7 2023

Cho: \(A=\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+....+\dfrac{2}{100^2}\)

\(A=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)\)

Và cho \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

Mà: 

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

....

\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)

Nên: \(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\)

\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow B< 1-\dfrac{1}{100}\)

\(\Rightarrow B< \dfrac{99}{100}\)

Mà: \(\dfrac{99}{100}< 1\) (tử nhỏ hơn mẫu)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

\(\Rightarrow A=2\cdot\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..+\dfrac{1}{100^2}\right)< 2\) (đpcm)

25 tháng 7 2023

\(\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+...+\dfrac{2}{100^2}\)

\(=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)\)

mà \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

\(\Rightarrow dpcm\)

4 tháng 2 2018
a, ko vt de bai =25*79-32*(25+47) =1975-32*72 =1975-2304 =-329 b, ko vt de bai =[-3×6]×(-5)×(-5) 18×25=4500 c, -9+27-0 =18-0 =18 bai2 a,
4 tháng 2 2018
b2 a, 4524-(-135)-4035 =4659-4035 =624 b, =100-709+(-280) =(-609)+(-280) = -889 c, =(-329)+(-86)-(-415) =(-415)-(-415) =0 k cho mik nha
15 tháng 2 2016

101 + 100 + ... + 2 + 1 = 101x102/2 = 101x51 = 5151 
101 - 100 + 99 - .. + 1 = ( 101 -100 ) + ( 99 - 98 ) + ... + ( 3 - 2 ) + 1 = 1 + 1 + 1 + ... + 1 ( 51 số ) = 51 
suy ra C = 5151/51 = 101 

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3737x43 - 4343x36 = 37x101x43 - 43x101x36 = 43x101 = 4343 
2 + 4 + 6 +... + 100 = 2x( 1 + 2 + ... + 50 ) = 2x50x51/2 = 50x51 = 2550 

vậy D = 4343/2550 

15 tháng 2 2016

b,D = 4343/2550