tính nhanh
(1+1/2).(1+1/3). ... .(1+1/100)
giúp mình với
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\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)
F = \(\frac{1}{2}\) . \(\frac{2}{3}\) ..... \(\frac{98}{99}\) .\(\frac{99}{100}\)
\(\Leftrightarrow\)F = \(\frac{1.2.3...98.99}{2.3.4...99.100}\)
\(\Leftrightarrow\)F = \(\frac{1}{100}\)
Vậy F =\(\frac{1}{100}\)
\(F=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{100}-1\right)\)
\(F=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{99}{100}\right)\)
F có : ( 99 - 1 ) : 1 + 1 = 99 phân số
=> F mang dấu âm
=> \(F=-\left(\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{99}{100}\right)\)
=> \(F=-\left(\frac{1\cdot2\cdot...\cdot99}{2\cdot3\cdot...\cdot100}\right)\)
=> \(F=-\frac{1}{100}\)
c/
C = 1/100-1/100-1/99-1/99-1/98-1/98-1/97-..........-1/3-1/2-1/2-1/1
C = 1/100-1/100-1/1
C = 0-1/1
C = -1
Ta có:\(3Q=3+3^2+3^3+............+3^{101}\)
\(\Rightarrow3Q-Q=\left(3+3^2+.......+3^{101}\right)-\left(1+3+......+3^{100}\right)\)
\(\Rightarrow2Q=3^{101}-1\Rightarrow Q=\frac{3^{101}-1}{2}\)
\(Q=1+3+3^2+...+3^{100}\)(1)
\(\Rightarrow3Q=3+3^2+3^3+...+3^{101}\)(2)
Lấy (2) trừ (1) ta có :
\(\Rightarrow2Q=3^{101}-1\)
\(\Rightarrow Q=\frac{3^{101}-1}{2}\)
Nhận xét :\(1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{\left(n+1\right)\left(n-1\right)}{n^2}\)
Do đó : \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{100^2}\right)\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.........\frac{99.101}{100^2}\)
\(=\frac{\left(1.2.3....99\right)\left(3.4.5....101\right)}{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}\)
\(=\frac{101}{100.2}\)
\(=\frac{101}{200}\)
A=1-2-3+4+5-6-7+8+...+97-98-99+100
=>A=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
=>A=0+0+....+0=0
vậy A=0
B=1-2+2^2-2^3+...+2^100
=>2B=2-2^2+2^3-2^4+....+2^101
=>2B+B=1-2^101=3B
=>B=1-2^101/3
C= 2^100-2^99-2^98-...-2^2-2-1
=>C=2^100-(2^99+2^98+.....+2^2+2+1)
Đặt D=2^99+2^98+.....+2^2+2+1
=>2D=2^100+2^99+.....+2^3+2^2+2
=>2D-D=2^100-1=D
=>C=2^100-(2^100-1)=1
tick nha
hic!ngày kia phải nộp rồi ! mọi người giúp mình nhanh nha!
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{100}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)...\left(\frac{100}{100}+\frac{1}{100}\right)\)
\(=\frac{3}{2}.\frac{4}{3}...\frac{101}{100}\)
\(=\frac{3.4...101}{2.3...100}\)
\(=\frac{101}{2}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{100}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)...\left(\frac{100}{100}+\frac{1}{100}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{101}{100}\)
\(=\frac{101}{2}\)