Cho S = \(\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{50^2}\) Chứng minh S không là số nguyên
S = \(\dfrac{1}{1^2}\) +\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+....+\(\dfrac{1}{50^2}\). Chứng minh rằng S < 2
1/2^2+1/3^2+...+1/50^2<1/1*2+1/2*3*+...+1/49*50
=1/1-1/2+1/2-1/3+...+1/49-1/50<1
=>S<1+1=2
\(S=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\) Chứng minh rằng \(S< \dfrac{1}{2}\)
\(S=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A< 1-\dfrac{1}{50}\Rightarrow A< 1\)
Ta có \(S=\dfrac{1}{2^2}\left(1+A\right)\)
Ta có
\(A< 1\Rightarrow1+A< 2\Rightarrow S< \dfrac{1}{2^2}.2=\dfrac{1}{2}\)
a)chứng minh rằng :\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)........+\(\dfrac{1}{100^2}< \dfrac{1}{2}\)
b)tính nhanh tổng S với S= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+......+\dfrac{1}{61.63}\)
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Cho: \(S=\dfrac{1^2-1}{1}+\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+....+\dfrac{n^2-1}{n^2}\)(n∈N*). CMR S không phải là số nguyên.
Lời giải:
$n=1$ thì $S=0$ nguyên nhé bạn. Phải là $n>1$
\(S=1-\frac{1}{1^2}+1-\frac{1}{2^2}+1-\frac{1}{3^2}+...+1-\frac{1}{n^2}\)
\(=n-\underbrace{\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)}_{M}\)
Để cm $S$ không nguyên ta cần chứng minh $M$ không nguyên. Thật vậy
\(M> 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}\)
\(M>1+\frac{1}{2}-\frac{1}{n+1}>1\) với mọi $n>1$
Mặt khác:
\(M< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{(n-1)n}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\)
\(M< 1+1-\frac{1}{n}< 2\)
Vậy $1< M< 2$ nên $M$ không nguyên. Kéo theo $S$ không nguyên.
a, Cho b là số tự nhiên, b>1. Chứng minh rằng: \(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b^2}< \dfrac{1}{b-1}-\dfrac{1}{b}\)
b, Áp dụng phần a: Cho S\(=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\). Chứng minh rằng: \(\dfrac{2}{5}< S< \dfrac{8}{9}\)
a)Ta có:\(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{b+1-b}{b\left(b+1\right)}=\dfrac{1}{b^2+b}< \dfrac{1}{b^2}\)(do b>1)
\(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{b-b+1}{\left(b-1\right)b}=\dfrac{1}{b^2-b}>\dfrac{1}{b^2}\)(do b>1)
b)Áp dụng từ câu a
=>\(\dfrac{1}{2}-\dfrac{1}{3}< \dfrac{1}{2^2}< \dfrac{1}{1}-\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{4}< \dfrac{1}{3^2}< \dfrac{1}{2}-\dfrac{1}{3}\)
.........................
\(\dfrac{1}{9}-\dfrac{1}{10}< \dfrac{1}{9^2}< \dfrac{1}{8}-\dfrac{1}{9}\)
=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}< S< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
=>\(\dfrac{1}{2}-\dfrac{1}{10}< S< 1-\dfrac{1}{9}\)
=>\(\dfrac{2}{5}< S< \dfrac{8}{9}\)(đpcm)
Chứng minh S<\(\dfrac{5}{32}\)
Cho \(S=\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{103^2}\)
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Cho \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}.\) Chứng minh rằng: \(S>\dfrac{9}{22}\)
Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
= 1/2-1/3+1/3-1/4+...+1/10-1/11
= 1/2 - 1/11 = 9/22 (đpcm)
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{2017}{5^{2017}}+\dfrac{2018}{5^{2018}}\).Chứng minh S<\(\dfrac{1}{3}\)
Cho S=\(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2023}{4^{2023}}\). Chứng minh S < \(\dfrac{1}{2}\)
=> 4S = 1 + 2/4 + 3/4^2 +...+ 2023/4^2022
=> 4S-S = 1 + (2/4-1/4) + (3/4^2 - 2/4^2) +...+ (2023/4^2022 - 2022/4^2022) - 2023/4^2023
=> 3S = 1 + 1/4 + 1/4^2 +...+ 1/4^2022 - 2023/4^2023
=> 12S = 4 + 1 + 1/4 +... + 1/4^2021 - 2023/4^2022
=> 12S - 3S = 4 + (1-1) + (1/4-1/4) +... + (1/4^2021 - 1/4^2021) - 1/4^2022 - 2023/4^2022 + 2023/4^2023
=> 9S = 4 - 1/4^2022 - 2023/4^2022 + 2023/4^2023
= 4- 2024/4^2022 + 2023/4^2023
Do 2024/4^2022 > 2024/4^2023 > 2023/4^2023 nên - 2024/4^2022 + 2023/4^2023 < 0
=> 9S < 4 < 9/2
=> S < 1/2 (đpcm)
Cho S=1+3+3^2+....+3^2023
Chứng tỏ S chia hết cho 4