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L
1
HT
2
H
1 tháng 3 2020
â,Đặt A là tên bthuc
A\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+2020\left(2021-1\right)\)
\(=\left(1.2+2.3+3.4+...+2020.2021\right)-\left(1+2+3+...+2020\right)\)
Đặt B = 1.2+2.3+...+2020.2021
\(3B=1.2.3+2.3.3+...+2020.2021.3\)
=\(1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+2020.2021.\left(2022-2019\right)\)
\(=\left(1.2.3+2.3.4+...+2020.2021.2022\right)-\left(0.1.2+1.2.3+...+2019.2020.2021\right)\)
\(=2020.2021.2022-0.1.2=2020.2021.2022\)
=>\(B=\frac{2020.2021.2022}{3}\)
=>\(A=\frac{2020.2021.2022}{3}-\frac{2020.2021}{2}=2020.2021\left(\frac{2022}{3}-\frac{1}{2}\right)=\frac{2020.2021.4041}{6}\)
H
1 tháng 3 2020
b,Đặt tên bthuc là M
Ta có: \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)
=> \(1^3-1=0.1.2\)
\(2^3=1.2.3\)
.......
\(2020^3-2020=2019.2020.2021\)
=> \(M=0.1.2+1.2.3+2.3.4+...+2019.2020.2021+\left(1+2+...2020\right)\)
Đặt N=1.2.3+2.3.4+...+2019.2020.2021
\(4N=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+2019.2020.2021\left(2022-2018\right)\)
=\(\left(1.2.3.4+2.3.4.5+...+2019.2020.2021.2022\right)\)-\(\left(0.1.2.3+1.2.3.4+...+2018.2019.2020.2021\right)\)
\(=2019.2020.2021.2022-0.1.2.3=2019.2020.2021.2022\)
=>\(N=\frac{2019.2020.2021.2022}{4}\)
=>\(M=\frac{2019.2020.2021.2022}{4}+\frac{2020.2021}{2}=\frac{2019.2020.2021.2022+2.2020.2021}{4}\)
\(=\frac{2020.2021\left(2019.2022+2\right)}{4}=\frac{2020.2021.\left(2019.2022-2019+2022-1\right)}{4}\)
\(=\frac{2020.2021.\left(2019+1\right)\left(2022-1\right)}{4}=\frac{2020^2.2021^2}{4}=\left(1010.2021\right)^2\)
TN
0
10 tháng 2 2017
\(a.\) \(\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.3^2.2^2+3^3}{-13}=\frac{2^3.3^3+3^3.2^2+3^3}{-13}\)
\(=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=\frac{3^3.\left(-1\right)}{1}=-27\)
\(b.\)\(A=2^2+4^2+6^2+...+20^2=2^2\left(1+2^2+3^2+...+10^2\right)\)
\(A=2^2.\frac{10.\left(10+1\right).\left(2.10+1\right)}{6}=4.385=1540\)
( Ta có: công thức tính tổng bình phương liên tiếp tứ 1 đến n là: \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\))
\(c.\)\(B=100^2+200^2+...+1000^2=\left(100.1\right)^2+\left(100.2\right)^2+...+\left(100.10\right)^2\)
\(B=100^2.1^2+100^2.2^2+...+100^2.10^2=100^2.\left(1^2+2^2+...+10^2\right)\)
Áp dụng công thức \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
Ta có: \(B=100^2\times385=3,850,000\)
N
13 tháng 8 2017
Bài 2 :
a, \(2^x+2^{x+4}=272\)
\(2^x+2^x.2^4=272\)
\(2^x.\left(1+2^4\right)=272\)
\(2^x.17=272\)
\(2^x=272:17\)
\(2^x=16=2^4\)
\(\Rightarrow x=4\)
13 tháng 8 2017
1: Bài này hơi khó đó
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6\times\left(6+1\right)\div2}+\frac{1}{7\times\left(7+1\right)\div2}+...+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6\times7\div2}+\frac{1}{7\times8\div2}+...+\frac{1}{x\times\left(x+1\right)\div2}\)
\(\Rightarrow\frac{2}{6\times7}+\frac{2}{7\times8}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\times\left(\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow2\times\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\div2\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
=> x = 18 - 1
=> x = 17
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dots+\dfrac{1}{2^{2020}}\)
\(2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dots+\dfrac{1}{2^{2019}}\)
\(2A-A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dots+\dfrac{1}{2^{2019}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dots+\dfrac{1}{2^{2020}}\right)\)
\(A=2-\dfrac{1}{2^{2020}}\)