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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
\(4N=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+2019.2020.2021\left(2022-2018\right)\)
=\(\left(1.2.3.4+2.3.4.5+...+2019.2020.2021.2022\right)\)-\(\left(0.1.2.3+1.2.3.4+...+2018.2019.2020.2021\right)\)
\(=2019.2020.2021.2022-0.1.2.3=2019.2020.2021.2022\)
=>\(N=\frac{2019.2020.2021.2022}{4}\)
=>\(M=\frac{2019.2020.2021.2022}{4}+\frac{2020.2021}{2}=\frac{2019.2020.2021.2022+2.2020.2021}{4}\)
\(=\frac{2020.2021\left(2019.2022+2\right)}{4}=\frac{2020.2021.\left(2019.2022-2019+2022-1\right)}{4}\)
\(=\frac{2020.2021.\left(2019+1\right)\left(2022-1\right)}{4}=\frac{2020^2.2021^2}{4}=\left(1010.2021\right)^2\)
TD
5
8 tháng 10 2020
Ta có: \(A=1+2+2^2+...+2^{2020}\)
\(\Leftrightarrow2A=2+2^2+2^3+...+2^{2021}\)
Trừ vế 2A cho A ta được:
\(2A-A=\left(2+2^2+...+2^{2021}\right)-\left(1+2+...+2^{2020}\right)\)
\(\Rightarrow A=2^{2021}-1\)
8 tháng 10 2020
Ta có: \(B=1+5+5^2+...+5^{2020}\)
\(\Leftrightarrow5B=5+5^2+5^3+...+5^{2021}\)
Trừ vế 5B cho B ta được:
\(5B-B=\left(5+5^2+...+5^{2021}\right)-\left(1+5+...+5^{2020}\right)\)
\(\Leftrightarrow4B=5^{2021}-1\)
\(\Rightarrow B=\frac{5^{2021}-1}{4}\)
TN
1
14 tháng 5 2020
Xét theo thừa số tổng quát:
1−1t2=t2−1t2=(t−1)(t+1)t21−1t2=t2−1t2=(t−1)(t+1)t2
Thay vào bài toán:
A=(1−122)(1−132)(1−142).....(1−1n2)
A=((2−1)(2+1)22)((3−1)(3+1)32)((4−1)(4+1)42)....((n−1)(n+1)n2)A=(1−122)(1−132)(1−142).....(1−1n2)=((2−1)(2+1)22)((3−1)(3+1)32)((4−1)(4+1)42)....((n−1)(n+1)n2)
A=1.32.2.2.43.3.3.54.4......(n−1)(n+1)n.n
A=1.2.3.....(n−1)2.3.4.....n.3.4.5......(n+1)2.3.4......n
=1n.n+12
=
\(\dfrac{n+1}{2n}\)
PT
0
PT
0