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\(E=-\dfrac{1}{3}\cdot\left(1+2+3\right)-\dfrac{1}{4}\left(1+2+3+4\right)-...-\dfrac{1}{50}\left(1+2+3+...+50\right)\)
\(=\dfrac{-1}{3}\cdot\dfrac{3\cdot4}{2}-\dfrac{1}{4}\cdot\dfrac{4\cdot5}{2}-...-\dfrac{1}{50}\cdot\dfrac{50\cdot51}{2}\)
\(=\dfrac{-4}{2}-\dfrac{5}{2}-...-\dfrac{51}{2}\)
\(=\dfrac{-\left(4+5+...+51\right)}{2}\)
\(=\dfrac{-\left(51+4\right)\cdot\dfrac{48}{2}}{2}=-\dfrac{1320}{2}=-660\)
mình chỉ biết câu b thôi:
Ta biến đổi vế phải :
1-1/2+1/3-1/4+.....+1/49-1/50
=(1+1/3+1/5+....+1/49)-(1/2+1/4+1/6+.......+1/50)
=(1+1/2+1/3+.....+1/49+1/50)-2(1/2+1/4+1/6+......+1/50)
=(1+1/2+...+1/50)-(1+1/2+1/3+....+1/25)
=1/26+1/27+.......+1/50
Vậy 1/26+1/27+1/28+.....+1/50=1-1/2+1/3-1/4+......+1/49-1/50
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};.......;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow3+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{50^2}< 1+3=4\)
Vậy \(3+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}< 4\)
Ta có : 1/2^2<1/1.2
1/3^2 < 1/2.3
1/4^2<1/3.4
................
.............
1/50^2<1/49.50
=> 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/50^2 < 1/1.2+1/2.3+1/3.4+....+1/49.50
=> 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/50^2 < 1-1/50
=> 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/50^2 < 49/50 < 1
Vậy 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/50^2 < 1