CHO \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{10^2}\)
tính S
Những câu hỏi liên quan
Bài 1 : Tính tổng S , biết : \(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{2010\times2011}\)
Bài 2 : Tính tổng sau : \(S=\frac{3}{10\times13}+\frac{3}{13\times16}+\frac{3}{16\times19}+....+\frac{3}{58\times61}\)
Bài 3 : Tính tổng sau : \(S=\frac{1}{4\times7}+\frac{1}{7\times10}+\frac{1}{10\times13}+....+\frac{1}{19\times22}\)
Bài 1 :
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
Bài 2 :
\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
Bài 3 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)
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Tính S = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
Cho S = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
Khi đó [S] = ....
Cho \(S=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}}+.......+\sqrt{1+\frac{1}{2^2}+...+\frac{1}{n^2}}\)
Viết quy trình bấm phím tính S
Tính S10, S12
Tính nhanh :a ) S 2+4+6+8+.....+2018b ) S 10+102+103+...+10100c) Sfrac{1}{5}+frac{1}{5^2}+frac{1}{5^3}+.....+frac{1}{5^{100}}d) Sfrac{1!}{3!}+frac{2!}{4!}+frac{3!}{5!}+....+frac{2018!}{2020!}biết rằng : n!1×2×3×...×nVD : 1! 12! 2.13!1.2.34!1.2.3.4! : giai thừa
Đọc tiếp
Tính nhanh :
a ) S = 2+4+6+8+.....+2018
b ) S= 10+102+103+...+10100
c) S=\(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+.....+\frac{1}{5^{100}}\)
d) S=\(\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+....+\frac{2018!}{2020!}\)
biết rằng : n!=1×2×3×...×n
VD : 1! = 1
2! = 2.1
3!=1.2.3
4!=1.2.3.4
! : giai thừa
a) 2 +4+6+8+...+2018
= ( 2018+2) x 1009 : 2
= 2020 x 1009 : 2
= 1009 x (2020:2)
= 1009 x 1010
= 1 019 090
b) S = 10 + 102 + 103 + ...+ 10100
=> 10.S = 102 + 103 + 104 +...+ 10101
=> 10.S - S = 10101-10
9.S=10101- 10
\(\Rightarrow S=\frac{10^{101}-10}{9}\)
c) \(S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(\Rightarrow5S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(5S-S=1-\frac{1}{5^{100}}\)
\(4S=1-\frac{1}{5^{100}}\)
\(S=\frac{1-\frac{1}{5^{100}}}{4}\)
e cx ko nx, e ms hok lp 7 thoy, sang hè ms lp 8! e sr cj nhiều nha!
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d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+...+\frac{2018!}{2020!}\)
\(S=\frac{1}{1.2.3}+\frac{1.2}{1.2.3.4}+\frac{1.2.3}{1.2.3.4.5}+...+\frac{1.2.3...2018}{1.2.3...2020}\)
\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2019.2020}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(S=\frac{1}{2}-\frac{1}{2020}\)
\(S=\frac{1009}{2020}\)
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Tính \(S=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2}+\frac{1}{3}\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+...+\left(1+\frac{1}{2}+...+\frac{1}{10}\right)\)
Các bạn giúp mình nhé !~~~
Cho \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\)
So sánh S với 1
Ta có
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..............
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
=> S < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
S < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(S< 1-\dfrac{1}{100}< 1\)(do 1/100 >0)
ĐPcm
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Giải:
\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{99^2}=\dfrac{1}{99.99}< \dfrac{1}{98.99}\)
\(\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{100}< 1\)
\(\Rightarrow S< 1\)
Vậy S < 1.
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tính S\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\)
S = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
2S = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
2S - S = \(1-\frac{1}{2^{100}}\)
=> S = \(1-\frac{1}{2^{100}}\)
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Xem thêm câu trả lời
Tính nhanh :a ) S 2+4+6+8+.....+2018b ) S 10+102+103+...+10100c) Sfrac{1}{5}+frac{1}{5^2}+frac{1}{5^3}+...+frac{1}{5^{100}}d) Sfrac{1!}{3!}+frac{2!}{4!}+frac{3!}{5!}+....+frac{2018!}{2020!}biết rằng : n!1times2times3times...times nVD : 1! 12! 2.13!1.2.34!1.2.3.4! : giai thừa
Đọc tiếp
Tính nhanh :
a ) S = 2+4+6+8+.....+2018
b ) S= 10+102+103+...+10100
c) \(S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+....+\frac{2018!}{2020!}\)
biết rằng : \(n!=1\times2\times3\times...\times n\)
VD : 1! = 1
2! = 2.1
3!=1.2.3
4!=1.2.3.4
! : giai thừa
a;b;c có những câu tương tự rồi, ko giải lại nx
d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+...+\frac{2018!}{2020!}\)
\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2019.2020}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(S=\frac{1}{2}-\frac{1}{2020}\)
b tự làm nốt nha
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