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Ta đặt
\(A=\dfrac{7}{1\times2}+\dfrac{7}{2\times3}+...+\dfrac{7}{99\times100}\)
\(\dfrac{1}{7}\times A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+....+\dfrac{1}{99\times100}\)
\(\dfrac{1}{7}\times A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\dfrac{1}{7}\times A=1-\dfrac{1}{100}\)
\(\dfrac{1}{7}\times A=\dfrac{99}{100}\)
\(A=\dfrac{99}{100}\div\dfrac{1}{7}\)
\(A=\dfrac{693}{100}\)
= 7.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100)
= 7.(1 - 1/100)
= 7 . 99/100
= 693/100
\(1.2^2+2.3^2+...+99.100^2\)
\(=1.2\left(3-1\right)+2.3\left(4-1\right)+...+99.100\left(101-1\right)\)
\(=1.2.3-1.2+2.3.4-2.3+...+99.100.101-99.100\)
\(=\left(1.2.3+2.3.4+...+99.100.101\right)\)\(-\left(1.2+2.3+...+99.100\right)\)
Chúc học tốt
Ta có : \(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\right)\)
\(\Rightarrow S=2.\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}=\frac{99}{50}\)
=2.(1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+.........+\(\frac{1}{99}\)-\(\frac{1}{100}\))
=2.(1-\(\frac{1}{100}\))
S= 2.\(\frac{99}{100}\)
S=\(\frac{99}{50}\)
Ta thấy mỗi tổng trên là tích của hai số tự nhiên liên tiếp.
\(a_1=1.2\Rightarrow3a_1=1.2.3\)\(\Rightarrow3a_1=1.2.3-0.1.2\).
\(a_2=2.3\Rightarrow3a_2=2.3.3\)\(\Rightarrow3a_2=2.3.4-1.2.3\).
.....
\(a_{99}=99.100\Rightarrow3a_{99}=3.99.100\)\(\Rightarrow3a_{99}=98.99.100-97.98.99\).
Ta có:
\(3A=1.2.3+2.3.3+3.4.3+....+99.100.3\)
\(=\)\(1.2.3-0.1.2+2.3.4-1.2.3+........+98.99.100-97.98.100\)
\(=98.99.100\)
Suy ra: \(A=\frac{98.99.100}{3}=323400\).
B=1.2+2.3+3.4+...+99.100
⇒3B=1.2.3+2.3.3+....+99.100.3
⇒3B=1.2.3+2.3.(4−1)+...+99.100.(101−98)
⇒3B=1.2.3+2.3.4−1.2.3+...+99.100.101−98.99.100
⇒3B=99.100.101
\(⇒\)
\(B=1.2+2.3+....+99.100\)
\(\Rightarrow3B=1.2.3+2.3.4+...+99.100.3\)
\(\Rightarrow3B=1.2.\left(3-0\right)+2.3.\left(4-1\right)+....+99.100.\left(101-98\right)\)
\(=\left(1.2.3+2.3.4+....+99.100.101\right)-\left(0.1.2+1.2.3+...+98.99.100\right)\)
\(=99.100.101-0.1.2\)
= 999900 - 0
=> B = 999900 : 3 = 333300
Vậy B = 333300
B = 1.2 + 2.3 + 3.4 + ...+ 99.100
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ...+99.100.3
3B = 1.2.3 + 2.3.(4-1) + ...+ 99.100.(101-98)
3B = 1.2.3 + 2.3.4 - 1.2.3 + ...+ 99.100.101 - 98.99.100
3B = (1.2.3+2.3.4+...+99.100.101) - (1.2.3+...+98.99.100)
3B = 99.100.101
\(\Rightarrow B=\frac{99.100.101}{3}=333300\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(B=1\cdot2^2+2\cdot3^2+3\cdot4^2+...+99\cdot100^2\\ =\left(2-1\right)\cdot2^2+\left(3-1\right)\cdot3^2+\left(4-1\right)\cdot4^2+...+\left(100-1\right)\cdot100^2\\ =2\cdot2^2-1\cdot2^2+3\cdot3^2-1\cdot3^2+4\cdot4^2-1\cdot4^2+...+100\cdot100^2-1\cdot100^2\\ =2^3-2^2+3^3-3^2+4^3-4^2+...+100^3-100^2\\ =1^3-1^2+2^3-2^2+3^3-3^2+4^3-4^2+...+100^3-100^2\\=\left(1^3+2^3+3^3+4^3+...+100^3\right)-\left(1^2+2^2+3^2+4^2+...+100^2\right)\\=\left(1+2+3+...+100\right)^2-\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\\ =\left[\dfrac{\left(100\cdot101\right)}{2}\right]^2-\dfrac{100\cdot101\cdot201}{6}\\ =5050^2-2030100\\ =25502500-2030100\\ =23472400 2\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
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