tinh:
1^3 + 2^3 + 3^3 + ... + 2020^3
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NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2020
\(\left(a+b+c\right)^2=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
\(\Rightarrow P=\frac{a^{2020}+1}{a^{2020}+a^{2020}+a^{2020}+3}=\frac{a^{2020}+1}{3\left(a^{2020}+1\right)}=\frac{1}{3}\)
19 tháng 9 2019
\(a,A=2^1+2^2+2^3+...+2^{2019}\)
\(2A=2^2+2^3+2^4+...+2^{2020}\)
\(\Rightarrow2A-A=A=2^{2020}-2\)
\(B=1+3+3^2+3^3+...+3^{2020}\)
\(3B=3+3^2+3^3+...+3^{2021}\)
\(3B-B=2B=3^{2021}-1\)
\(B=\frac{3^{2021}-1}{2}\)
19 tháng 9 2019
a,\(A=2^1+2^2+2^3+...+2^{2019}\)
\(2A=2^2+2^3+2^4+...+2^{2020}\)
\(2A-A=\left[2^2+2^3+2^4+...+2^{2020}\right]-\left[2^1+2^2+...+2^{2019}\right]\)
\(A=2^{2020}-2^1=2^{2020}-2\)
b, \(B=1+3+3^2+3^3+...+3^{2020}\)
\(3B=3+3^2+3^3+...+3^{2021}\)
\(3B-B=\left[3+3^2+3^3+...+3^{2021}\right]-\left[1+3+3^2+...+3^{2020}\right]\)
\(2B=3^{2021}-1\)
\(B=\frac{3^{2021}-1}{2}\)
PL
1
3 tháng 4 2019
\(B=\frac{2019}{1}+\frac{2018}{2}+\frac{2017}{3}+......+\frac{1}{2019}\)
\(=\left(\frac{2018}{2}+1\right)+\left(\frac{2017}{3}+1\right)+.....+\left(\frac{1}{2019}+1\right)+1\)
\(=\frac{2020}{2}+\frac{2020}{3}+\frac{2020}{4}+.....+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2020}\right)\)
\(=2020A\)
\(\Rightarrow\frac{A}{B}=\frac{A}{2020A}=\frac{1}{2020}\)
D
2
12 tháng 3 2020
Đặt \(A=1-3^2+3^3-3^4+...+3^{2017}-3^{2018}+3^{2019}-3^{2020}\)
\(\Leftrightarrow A=1-\left(3^2-3^3+3^4-.....-3^{2017}+3^{2018}-3^{2019}+3^{2020}\right)\)
Đặt \(B=3^2-3^3+3^4-.....-3^{2017}+3^{2018}-3^{2019}+3^{2020}\)
\(3B=3\left(3^2-3^3+3^4-.....-3^{2017}+3^{2018}-3^{2019}+3^{2020}\right)\)
\(3B=3^3-3^4+3^5-....-3^{2018}+3^{2019}-3^{2020}+3^{2021}\)
\(3B+B=\left(3^3-3^4+3^5-....-3^{2018}+3^{2019}-3^{2020}+3^{2021}\right)\)
\(+\left(3^2-3^3+3^4-.....-3^{2017}+3^{2018}-3^{2019}+3^{2020}\right)\)
\(4B=3^{2021}+3^2\)
\(B=\frac{3^{2021}+3^2}{4}\)Thay vào A ta có A=\(1-\frac{3^{2021}+3^2}{4}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
2 tháng 1 2019
\(M=2.3.4.4+3.4.5.4+...+2018.2019.2020.4\)
\(M=2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+2018.2019.2020.\left(2021-2017\right)\)
M=2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+2018.2019.2020.2021-2017.2018.2019.2020=2018.2019.2020-1.2.3.4
VD
0
VD
0
12 tháng 3 2020
Đặt \(A=5+5^2+5^3+....+5^{199}+5^{200}\)
\(\Leftrightarrow5A=5\left(5+5^2+5^3+....+5^{199}+5^{200}\right)\)
\(\Leftrightarrow5A=5^2+5^3+5^4+....+5^{200}+5^{201}\)
\(\Leftrightarrow5A-A=\left(5^2+5^3+5^4+....+5^{200}+5^{201}\right)-\left(5+5^2+5^3+....+5^{199}+5^{200}\right)\)
\(\Leftrightarrow4A=5^{201}-5\)
\(\Leftrightarrow A=\frac{5^{201}-5}{4}\)
trên mạng cũng có mà bnaj