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\(3A=1.2.3+2.3.\left(4-1\right)+......+49.50.\left(51-48\right)\)
\(3A=1.2.3+2.3.4-1.2.3+........+49.50.51-48.49.50\)
\(3A=49.50.51\)
\(\Rightarrow A=\frac{49.50.51}{3}\)
\(\Rightarrow A=41650\)
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Lời giải:
\(A=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-998}{999}.\frac{-999}{1000}\\
=\frac{(-1)(-2)(-3)...(-998)(-999)}{2.3.4....1000}\\
=-\frac{1.2.3.4....998.999}{2.3.4...1000}\\
=-\frac{1}{1000}\)
Trong $B$ có một thừa số là $1-\frac{7}{7}=0$ nên $B=0$ (do số nào nhân với $0$ cũng sẽ bằng $0$.
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$C=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{49.51}{50^2}$
$=\frac{1.3.2.4.3.5.....49.51}{2^2.3^2.4^2....50^2}$
$=\frac{(1.2.3...49)(3.4.5...51)}{(2.3.4...50)(2.3.4...50)}$
$=\frac{1.2.3...49}{2.3.4...50}.\frac{3.4.5...51}{2.3.4....50}$
$=\frac{1}{50}.\frac{51}{2}=\frac{51}{100}$
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A) \(A=1+4+4^2+4^3+...+4^{26}\)
\(\Rightarrow4A=4+4^2+4^3+4^4+...+4^{27}\)
\(\Rightarrow4A-A=4^{27}-1\)
\(3A=4^{27}-1\)
\(A=\frac{4^{27}-1}{3}\)
B) \(B=3+3^3+3^5+3^7+...+3^{21}\)
\(\Rightarrow3^2B=3^3+3^5+3^7+3^9+...+3^{23}\)
\(\Rightarrow3^2B-B=3^{23}-3\)
\(8B=3^{23}-3\)
\(B=\frac{3^{23}-3}{8}\)
C) \(M=1.2+2.3+3.4+...+49.50\)
\(\Rightarrow3M=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(3M=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(3M=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(3M=\left(1.2.3+2.3.4+3.4.5+...+49.50.51\right)-\left(1.2.3+3.4.5+...+48.49.50\right)\)
\(3M=49.50.51\)
\(M=\left(49.50.51\right):3\)
\(M=41650\)
D) \(N=1.2.3+2.3.4+3.4.5+...+49.50.51\)
\(\Rightarrow4N=1.2.3.4+2.3.4.4+3.4.5.4+...+49.50.51.4\)
\(4N=1.2.3.\left(4-0\right)+2.3.4\left(5-1\right)+3.4.5.\left(6-2\right)+...+49.50.51.\left(52-48\right)\)
RỒI BN LÀM GIỐNG NHƯ MK Ở PHẦN C THÌ NÓ SẼ RA!
CHÚC BN HỌC TỐT!!!!
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\(a=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
\(a=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)
\(\Rightarrow a=\frac{100}{2}\)
\(\Rightarrow a=50\)
Xét \(a^2-1\)
Thay a=50 ta có :
\(50^2-1\)
\(=2500-1\)
\(=2499\)
Ta có :\(B=49.50\)
\(\Rightarrow B=2450\)
Mà \(2450< 2499\)
\(\Rightarrow a>B\)
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http://olm.vn/hoi-dap/question/157302.html
\(\text{Đ}\text{ặt}\)\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{49}+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{50}-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\)
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vậy thì tổng của : -1+(-2)+(-3)+.........+(-49) = -(1+2+3+..........+49) = -1225
\(D=1\cdot2+2\cdot3+3\cdot4+...+49\cdot50\)
\(3D=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+49\cdot50\cdot3\)
\(3D=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+49\cdot50\cdot\left(51-48\right)\)
\(3D=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+49\cdot50\cdot51-48\cdot49\cdot50\)
\(3D=49\cdot50\cdot51\)
\(D=\frac{49\cdot50\cdot51}{3}=41650\)