tinh :
1^2 + 2^2 +...+2019^2
K
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PL
1
3 tháng 4 2019
\(B=\frac{2019}{1}+\frac{2018}{2}+\frac{2017}{3}+......+\frac{1}{2019}\)
\(=\left(\frac{2018}{2}+1\right)+\left(\frac{2017}{3}+1\right)+.....+\left(\frac{1}{2019}+1\right)+1\)
\(=\frac{2020}{2}+\frac{2020}{3}+\frac{2020}{4}+.....+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2020}\right)\)
\(=2020A\)
\(\Rightarrow\frac{A}{B}=\frac{A}{2020A}=\frac{1}{2020}\)
7 tháng 12 2018
\(2A=2-2^2+2^3-...+2^{2019}\)
\(\Rightarrow2A+A=3A=2^{2019}+1\)
\(\Rightarrow3A-2^{2019}=2^{2019}+1-2^{2019}=1\)
8 tháng 2 2020
\(x^2=yz\Rightarrow\frac{x}{y}=\frac{z}{x}\left(1\right)\)
\(y^2=xz\Rightarrow\frac{x}{y}=\frac{y}{z}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\frac{x}{y}=\frac{y}{z}=\frac{z}{x}=\frac{x+y+z}{y+z+x}=1\)
\(\Rightarrow x=y=z\)
Thay y, z bằng x \(\Rightarrow M=\frac{3.x^{2019}}{\left(3x\right)^{2019}}=\frac{3x^{2019}}{3^{2019}.x^{2019}}=\frac{1}{3^{2018}}\)
X
20 tháng 1 2019
\(P=1+3^2+3^5+...+3^{2019}\)
\(\Rightarrow3^3P=3^3+3^5+3^8+...+3^{2022}\)
\(\Rightarrow3^3P-P=\left(3^3+3^5+3^8+...+3^{2022}\right)-\left(1+3^2+3^5+...+3^{2019}\right)\)
\(\Rightarrow P\left(3^3-1\right)=3^3+3^{2022}-1-3^2\)
\(\Rightarrow26P=3^{2022}+17\)
\(\Rightarrow P=\frac{3^{2022}+17}{26}\)
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, \(B=1+3+3^2+3^3+...+3^{2020}\)
\(3B=3+3^2+3^3+...+3^{2021}\)
\(3B-B=\left[3+3^2+3^3+...+3^{2021}\right]-\left[1+3+3^2+...+3^{2020}\right]\)
\(2B=3^{2021}-1\)
\(B=\frac{3^{2021}-1}{2}\)
HG
2
XO
25 tháng 7 2019
B = 1 + 52 + 53 + ... + 52019
5B = 5 + 52 + 53 + ... + 52020
Lấy 5B trừ B theo vế ta có :
5B - B = (5 + 52 + 53 + ... + 52020) - (1 + 52 + 53 + ... + 52019)
4B = 52020 - 1
B = \(\frac{5^{2020}-1}{4}\)
Vậy B = \(\frac{5^{2020}-1}{4}\)
25 tháng 7 2019
\(B=1+5+5^2+5^3+...+5^{2019}\)
\(\Leftrightarrow5B=5+5^2+5^3+...+5^{2020}\)
\(\Leftrightarrow5B-B=\left(5+5^2+...+5^{2020}\right)-\left(1+5+5^2+...+5^{2019}\right)\)
\(\Leftrightarrow4B=5^{2020}-1\Leftrightarrow B=\frac{5^{2020}-1}{4}\)
LA
3 tháng 3 2019
Không tính thì sao mà làm được :)
a)
\(2020-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2019^2}\)
\(=3+\left(1-\dfrac{1}{3^2}\right)+\left(1-\dfrac{1}{4^2}\right)+....+\left(1-\dfrac{1}{2019^2}\right)\)
\(=3+\left(\dfrac{3^2-1}{3^2}+\dfrac{4^2-1}{4^2}+...+\dfrac{2019^2-1}{2019^2}\right)\)
\(=3+\left(\dfrac{2\cdot4}{3^2}+\dfrac{3\cdot5}{4^2}+\dfrac{4\cdot6}{5^2}+\dfrac{5\cdot7}{6^2}+...+\dfrac{2018\cdot2020}{2019^2}\right)\)
\(=3+\dfrac{\left(2\cdot3\cdot4\cdot....\cdot2018\right)}{3\cdot4\cdot5\cdot6...\cdot2019}\cdot\dfrac{\left(3\cdot4\cdot5\cdot....\cdot2020\right)}{3\cdot4\cdot5\cdot6\cdot....\cdot2019}=3+\dfrac{2\cdot2020}{2019}\)
\(=\dfrac{10097}{2019}\)
N
3 tháng 3 2019
Có: \(\dfrac{1}{k^2}=\dfrac{1}{k.k}< \dfrac{1}{\left(k-1\right)k}\left(k\in\text{ℕ},k>0\right)\)
\(\Rightarrow A=2020-\dfrac{1}{3^2}-\dfrac{1}{4^2}-\dfrac{1}{5^2}-...-\dfrac{1}{2019^2}\)
\(A=2020-\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{2019^2}\right)\)
\(>2020-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}\right)\)
Có: \(\dfrac{1}{k-1}-\dfrac{1}{k}=\dfrac{1}{k\left(k-1\right)}\left(k\in\text{ℕ},k>0\right)\)
\(\Rightarrow A>2020-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{2018}-\dfrac{1}{2019}\right)\)
\(A>2020-\dfrac{1}{2}+\dfrac{1}{2019}\)>2,2
Có: \(B=\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{17}\)
\(B=\dfrac{1}{5}+\left(\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\right)\)\(< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{6}+...+\dfrac{1}{6}\)
\(=\dfrac{1}{5}+\dfrac{1}{6}.12=2+\dfrac{1}{5}=2,2\)
Vậy A>B.
NC
1
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