DP
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NH
0
13 tháng 11 2023
1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{97}\right)\)
\(=30\left(1+2^4+...+2^{96}\right)⋮30\)
2:
\(B=3+3^2+3^3+...+3^{2022}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)
\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)
ND
1
24 tháng 4 2021
Ta thấy:
\(2^2=2.2>1.2\Rightarrow\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(3^2=3.3>2.3\Rightarrow\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.................
\(9^2=9.9>8.9\Rightarrow\dfrac{1}{9^2}< \dfrac{1}{8.9}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\)
=> Đpcm
EK
8 tháng 5 2021
Ta thấy:
22=2.2>1.2⇒122<11.222=2.2>1.2⇒122<11.2
32=3.3>2.3⇒132<12.332=3.3>2.3⇒132<12.3
.................
92=9.9>8.9⇒192<18.992=9.9>8.9⇒192<18.9
⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9
⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89
=> ...(tự viết)
Ta thấy:
22=2.2>1.2⇒122<11.222=2.2>1.2⇒122<11.2
32=3.3>2.3⇒132<12.332=3.3>2.3⇒132<12.3
.................
92=9.9>8.9⇒192<18.992=9.9>8.9⇒192<18.9
⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9
⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89
=> 11111111111111111111110101010110000
HACK
NB
0
D
1
L
28 tháng 4 2022
Đặt A=11⋅2+12⋅3+...+17⋅8A=11⋅2+12⋅3+...+17⋅8
Dễ thấy: B=122+132+...+182B=122+132+...+182<A=11⋅2+12⋅3+...+17⋅8(1)<A=11⋅2+12⋅3+...+17⋅8(1)
Ta có:A=11⋅2+12⋅3+...+17⋅8A=11⋅2+12⋅3+...+17⋅8
=1−12+12−13+...+17−18=1−12+12−13+...+17−18
=1−18<1(2)=1−18<1(2)
Từ (1);(2)(1);(2) ta có: B<A<1⇒B<1
Gọi biểu thức này là \(A\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{302^2}\)
Với mọi số tự nhiên ta luôn có : \(\left(n-1\right)^2< n^2< \left(n+1\right)^2\)
\(\Rightarrow\frac{1}{\left(n+1\right)^2}< \frac{1}{n^2}< \frac{1}{\left(n-1\right)^2}\)
Áp dụng ta có :
\(\Rightarrow A< \frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+......+\frac{1}{30^2}\)
\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{29.30}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{29}-\frac{1}{30}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{30}< \frac{1}{2}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
TA CÓ:
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{30^2}< \frac{1}{29.30}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{29.30}\)
ta lại có:\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{29.30}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{30-29}{29.30}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{29}-\frac{1}{30}\)
\(=1-\frac{1}{30}< 1\)
mà \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{29.30}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}< 1-\frac{1}{30}< 1\left(đpcm\right)\)