chứng minh B=1/5^2+1/6^2+1/7^2+...+1/2020^2>1/5
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\(\dfrac{2}{5^2}=\dfrac{2}{5.5}< \dfrac{2}{4.5}\\\dfrac{2}{6^2}=\dfrac{2}{6.6}< \dfrac{2}{5.6}\)
Làm tương tự với những số hạng còn lại
Khi đó:
\(A=\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{2}{2020^2}\\ < \dfrac{2}{4.5}+\dfrac{2}{5.6}+...+\dfrac{2}{2019.2020}\\ =2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\right)\\ =2\left(\dfrac{1}{4}-\dfrac{1}{2020}\right)=\dfrac{252}{505}< \dfrac{252}{504}=\dfrac{1}{2}\left(đpcm\right)\)
Chúc em học tốt!
Bài 1:
$-1+2-3+4-5+6-7+8-...-2019+2020-2021$
$=(2+4+6+8+...+2020)-(1+3+5+...+2021)$
$=(\frac{2020-2}{2}+1).\frac{2020+2}{2}-(\frac{2021-1}{2}+1).\frac{2021+1}{2}=1021110- 1022121=-1011$
Bài 1 cách 2:
$A=-1+2-3+4-5+6-7+8-....-2019+2020-2021$
$=-1+(2-3)+(4-5)+(6-7)+....+(2020-2021)$
$=-1+\underbrace{(-1)+(-1)+...+(-1)}_{1010}=-1+(-1).1010=-1011$
Bài 3:
\(A=5+5^2+..+5^{12}\)
\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)
\(5A=5^2+5^3+...+5^{13}\)
\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)
\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)
\(4A=5^{13}-5\)
\(A=\dfrac{5^{13}-5}{4}\)
\(S=\left(1+5^2+5^4+5^6\right)+...+\left(5^{2014}+5^{2016}+5^{2018}+5^{2020}\right)\\ S=\left(1+5^2+5^4+5^6\right)+...+5^{2014}\left(1+5^2+5^4+5^6\right)\\ S=\left(1+5^2+5^4+5^6\right)\left(1+...+5^{2014}\right)\\ S=16276\left(1+...+5^{2014}\right)⋮313\left(16276⋮313\right)\)
Ta có \(n^2+\left(n+1\right)^2>2n\left(n+1\right)\)
=>\(\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2}\left(\frac{1}{n\left(n+1\right)}\right)=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
Áp dụng ta có \(\frac{1}{5}=\frac{1}{1^2+2^2}< \frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}\right)\)
\(\frac{1}{13}=\frac{1}{2^2+3^2}< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)\)
..................................................................
\(\frac{1}{2019^2+2020^2}< \frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)
=> \(VT< \frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\right)=\frac{1}{2}\left(1-\frac{1}{2020}\right)< \frac{1}{2}\)(ĐPCM)
Answer:
\(S=\left(1+5^2+5^4+5^6\right)+...+\left(5^{2014}+5^{2016}+5^{2018}+5^{2020}\right)\)
\(=\left(1+5^2+5^4+5^6\right)+...+5^{2014}+\left(1+5^2+5^4+5^6\right)\)
\(=\left(1+5^2+5^4+5^6\right).\left(1+...+5^{2014}\right)\)
\(=16276.\left(1+5^2+...+5^{2014}\right)⋮313\)
Mà ta có: \(S=16276⋮313\)
Vậy \(S⋮313\)