1/1+2+ 1/1+2+3 +1/1+2+3+4+...+1/1+2+3+4+...+50
E=-1/3.(1+2+3)-1/4.(1+2+3+4)-...-1/50.(1+2+3+4+...+50)
\(E=-\dfrac{1}{3}\cdot\left(1+2+3\right)-\dfrac{1}{4}\left(1+2+3+4\right)-...-\dfrac{1}{50}\left(1+2+3+...+50\right)\)
\(=\dfrac{-1}{3}\cdot\dfrac{3\cdot4}{2}-\dfrac{1}{4}\cdot\dfrac{4\cdot5}{2}-...-\dfrac{1}{50}\cdot\dfrac{50\cdot51}{2}\)
\(=\dfrac{-4}{2}-\dfrac{5}{2}-...-\dfrac{51}{2}\)
\(=\dfrac{-\left(4+5+...+51\right)}{2}\)
\(=\dfrac{-\left(51+4\right)\cdot\dfrac{48}{2}}{2}=-\dfrac{1320}{2}=-660\)
Thực Hiện phép tính
B= 1 + 1/2(1+2) + 1/3 (1+2+3)+1/4(1+2+3+4)+....+1/50(1+2+3+....+50)
\(B=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{50\cdot51}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{51}{2}\)
\(=\dfrac{50\cdot\dfrac{\left(51+2\right)}{2}}{2}=50\cdot\dfrac{53}{4}=662.5\)
E=-1/3.(1+2+3)-1/4.(1+2+3+4)-...-1/50.(1+2+3+4+...+50)
Giúp mình với
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+1/(1+2+3+4+5)...+1/(1+2+3+4+5...+99)+1/50 là:
Đặt \(S=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+.....+99}+\frac{1}{50}\)
Đặt E = \(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+....+99}\)
\(E=\frac{1}{2.3:2}+\frac{1}{3.4:2}+....+\frac{1}{99.100:2}\)
\(\frac{1}{2}E=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
E = 49/100 : 1/2 = 49/50
Vậy \(S=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)
1/2*1/2+1/3*1/3+1/4*1/4+....+1/50*1/50
Tính tổng :\(S=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+....+\frac{1}{50}.\left(1+2+3+4+....+50\right)\)
P= 1 + 1/1+2 + 1/1+2+3 + 1/1+2+3+4 + ... + 1/ 1+2+3+4+...+50
\(P=1+\frac{1}{1+2}+\frac{1}{1+2+3}+........+\frac{1}{1+2+3+.......+50}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+......+\frac{1}{\frac{50.51}{2}}=1+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{50.51}=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{50.51}\right)\) \(Taco:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\)
\(\Rightarrow P=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.......+\frac{1}{50}-\frac{1}{51}\right)=1+2\left(\frac{1}{2}-\frac{1}{51}\right)=1+1-\frac{2}{51}=2-\frac{2}{51}=\frac{100}{51}\)
#)Giải :
Đặt \(A=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+50}\)
\(\Rightarrow A=\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{50.50:2}\)
\(\Rightarrow A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{100.101}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{51}\)
\(\Rightarrow\frac{1}{2}A=\frac{49}{102}\)
\(\Rightarrow A=\frac{49}{102}.\frac{1}{2}\)
\(\Rightarrow A=\frac{49}{204}\)
1:
[1/1+2]+[1/1+2+3]+[1/1+2+3+4]+...+[1/1+2+3+4+...+50
tinh 1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+...+99)+1/50
sao bn ko tra trên mạng ấy