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Cô làm rồi em nhá
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Câu a, xem lại đề bài
Câu b:
P = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ...+ \(\dfrac{1}{2023^2}\)
Vì \(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\) = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)
\(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)
\(\dfrac{1}{4^2}\) < \(\dfrac{1}{3.4}\) = \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)
........................
\(\dfrac{1}{2023^2}\) < \(\dfrac{1}{2022.2023}\) = \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\)
Cộng vế với vế ta có:
0< P < 1 - \(\dfrac{1}{2023}\) < 1
Vậy 0 < P < 1 nên P không phải là số tự nhiên vì không tồn tại số tự nhiên giữa hai số tự nhiên liên tiếp
Câu c:
C = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{5^2}\) + \(\dfrac{1}{7^2}\) + ....+ \(\dfrac{1}{2021^2}\) + \(\dfrac{1}{2023^2}\) = C
B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+ \(\dfrac{1}{2020^2}\) + \(\dfrac{1}{2023^2}\) > 0
Cộng vế với vế ta có:
C+B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\)+ \(\dfrac{1}{6^2}\)+...+ \(\dfrac{1}{2023^2}\) > C + 0 = C > 0
Mặt khác ta có:
1 > \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{2023^2}\) (cm ở ý b)
Vậy 1 > C > 0 hay C không phải là số tự nhiên (đpcm)
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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 * .
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Giải:
A=1/22+1/32+1/42+...+1/92
Ta có:
1/22<1/1.2
1/32<1/2.3
1/42<1/3.4
...
1/92<1/8.9
⇒A<1/1.2+1/2.3+1/3.4+...+1/8.9
A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9
A<1/1-1/9
A<8/9
Ta có:
1/22>1/2.3
1/32>1/3.4
1/42>1/4.5
...
1/92>1/9.10
⇒A>1/2.3+1/3.4+1/4.5+...+1/9.10
A>1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
A>1/2-1/10
A>2/5
Vậy 2/5<A<8/9 (đpcm)
Chúc bạn học tốt!
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)
\(\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+...+\dfrac{2}{100^2}\)
\(=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)\)
mà \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
\(\Rightarrow dpcm\)