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B= 333300
C=328350
D=(n+1) /( n nhân 2)
E=(1/3 trừ 1/3^100):2
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
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
Ta có: \(1-\frac{2}{n.\left(n+1\right)}\)
=\(\frac{n.\left(n+1\right)-2}{n\left(n+1\right)}\)
=\(\frac{n^2+n-2}{n.\left(n+1\right)}\)
=\(\frac{\left(n^2-1\right)+\left(n-1\right)}{n.\left(n+1\right)}\)
=\(\frac{\left(n-1\right).\left(n+1\right)+\left(n-1\right)}{n.\left(n+1\right)}\)
=\(\frac{\left(n-1\right).\left(n+1+1\right)}{n.\left(n+1\right)}\)
=\(\frac{\left(n-1\right).\left(n+2\right)}{n.\left(n+1\right)}\)
=>\(1-\frac{2}{n.\left(n+1\right)}=\frac{\left(n-1\right).\left(n+2\right)}{n.\left(n+1\right)}\left(1\right)\)
Lại có: \(M=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)....\left(1-\frac{2}{99.100}\right)\)
=> \(M=\left(1-\frac{2}{2.\left(2+1\right)}\right).\left(1-\frac{2}{3.\left(3+1\right)}\right).\left(1-\frac{2}{4.\left(4+1\right)}\right)....\left(1-\frac{2}{99.\left(99+1\right)}\right)\left(2\right)\)
Thay (1) vào (2) ta được:
\(M=\frac{\left(2-1\right).\left(2+2\right)}{2.\left(2+1\right)}.\frac{\left(3-1\right).\left(3+2\right)}{3.\left(3+1\right)}.\frac{\left(4-1\right).\left(4+2\right)}{4.\left(4+1\right)}...\frac{\left(99-1\right).\left(99+2\right)}{99.\left(99+1\right)}\)
=> \(M=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{98.101}{99.100}\)
=> \(M=\frac{1.4.2.5.3.6....98.101}{2.3.3.4.4.5....99.100}\)
=> \(M=\frac{\left(1.2.3....98\right).\left(4.5.6....101\right)}{\left(2.3.4....99\right).\left(3.4.5....100\right)}\)
=> \(M=\frac{1.101}{99.3}\)
=> \(M=\frac{101}{297}\)
Vậy \(M=\frac{101}{297}\)
Gọi tổng trên là A
A=1/1.2.3+1/2.3.4+1/3.4.5+...1/98.99.100
Ta xét :
1/1.2 ‐ 1/2.3 = 2/1.2.3; 1/2.3 ‐ 1/3.4 = 2/2.3.4;...; 1/98.99 ‐ 1/99.100 = 2/98.99.100
tổng quát: 1/n﴾n+1﴿ ‐ 1/﴾n+1﴿﴾n+2﴿ = 2/n﴾n+1﴿﴾n+2﴿.
Do đó: 2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/98.99.100
= ﴾1/1.2 ‐ 1/2.3﴿ + ﴾1/2.3 ‐ 1/3.4﴿ +...+ ﴾1/98.99 ‐ 1/99.100﴿
= 1/1.2 ‐ 1/2.3 + 1/2.3 ‐ 1/3.4 + ... + 1/98.99 ‐ 1/99.100
= 1/1.2 ‐ 1/99.100
= 1/2 ‐ 1/9900
= 4950/9900 ‐ 1/9900
= 4949/9900.
Vậy A = 4949 / 9900
Bn làm sai r . kết quả là \(\frac{101}{297}\) nhưng mik ko bt cách giải thôi
=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)