a) Tính \(\frac{1}{n}-\frac{1}{n+1}\left(n\inℕ^∗\right)\)
b)Tính \(\frac{1}{n}-\frac{1}{n+3}\left(n\inℕ^∗\right)\)
c) Tính \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{3}{91.94}\)
d) Tính \(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}\)
Những câu hỏi liên quan
Chứng minh rằng với mọi n \(\inℕ^∗\):
D = \(\frac{1}{1.2}\frac{1}{2.3}\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}< 1\)
F = \(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n\left(n+2\right)}\right)< 2\)
So sánh:
a) \(A=\frac{n}{n+1};B=\frac{n+2}{n+3}\left(n\inℕ\right)\)
b) \(A=\frac{n}{n+3};B=\frac{n-1}{n+4}\left(n\inℕ^∗\right)\)
c) \(A=\frac{n}{2n+1};B=\frac{3n+1}{6n+3}\left(n\inℕ\right)\)
Giúp mình nhé gấp lắm ai trả lời đầu tiên mình sẽ tick
a)A=n/n+1=n/n+0/1
B=n+2/n+3=n/n + 2/3
ta có:0<2/3
=>A<B
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Với \(n\inℕ^∗\), cho:
\(A=1+\frac{1}{3}+...+\frac{1}{2n-3}+\frac{1}{2n-1}\)
\(B=\frac{1}{1\left(n-1\right)}+\frac{1}{3\left(2n-3\right)}+...+\frac{1}{\left(2n-3\right)\cdot3}+\frac{1}{\left(2n-1\right)\cdot1}\)
Tính \(\frac{A}{B}\).
1.Cho E=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+n}\right)\)và F=\(\frac{n+2}{n}\)\(\forall\)\(n\inℕ^∗\)Tính \(\frac{E}{F}\)
\(n\ge3;n\inℕ\)
CMR:
\(\frac{1}{a^n\left(b+c\right)}+\frac{1}{b^n\left(c+a\right)}+\frac{1}{c^n\left(a+b\right)}\ge\frac{3}{2}\)
chứng minh rằng :
a) \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\) ( n , a ϵ N* )
b) áp dụng câu a tính ;
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
a) \(\frac{1}{n}-\frac{1}{n+a}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{a\left(n+a\right)}\) (đpcm)
b) \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)
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a. Chứng minh rằng
\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\left(n,a\in Nsao\right)\)
b. Áp dụng câu a tính:
A= \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
B= \(\frac{5}{1.4}+\frac{5}{4.7}+..+\frac{5}{100.103}\)
C= \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
b) A=1/2.3+1/3.4+....+1/99.100
=> A=1/2-1/3+1/3-1/4+....+1/99-1/100
=> A=1/2-1/100
=> A=50/100-1/100
=> A=49/100
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Xem thêm câu trả lời
1.Tính:
a,\(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-......-\frac{1}{\left(n-1\right).n}\)\(n\in N\)
b,\(\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-....-\frac{4}{\left(n-4\right).n}\)\(n\in N\)
c\(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-.....-\frac{1}{2^{10}}\)
1 Tính :
a) \(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{n}\)
\(=\frac{1}{n}\)
b) \(B=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)
\(=\frac{4}{1.5}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\frac{1}{5}+\frac{1}{n}\)
\(=\frac{3}{5}+\frac{1}{n}\)
c) \(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow C=1-B\left(1\right)\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
Lấy 2B trừ B ta có :
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(B=1-\frac{1}{2^{10}}\left(2\right)\)
Thay (2) vào (1) ta có :
\(C=1-\left(1-\frac{1}{10}\right)\)
\(=1-1+\frac{1}{10}\)
\(=\frac{1}{10}\)
Vậy \(C=\frac{1}{10}\)
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Tính tổng sau
1) B= 1.2+2.3+3.4+......+99.100
2) C= \(1^2+2^2+3^2+...+99^2\)
3) D= \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{n^2}\right)\)
4) E=\(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{100}}\)
B= 333300
C=328350
D=(n+1) /( n nhân 2)
E=(1/3 trừ 1/3^100):2
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1)=>3B=1.2.3+2.3.3+3.4.3+...+99.100.3
3B=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
3B=99.100.101
=>B=333300
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