P= 1 + 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
1/1+2+ 1/1+2+3 +1/1+2+3+4+...+1/1+2+3+4+...+50
=2/6+2/12+....+2/51.50
=2(1/2.3+1/3.4+...+1/51.50)
=2(1/2 - 1/3 + ....... + 1/50 -1/51)
=2(1/2-1/51)
=2.49/102=49/51
học tốt
1:
[1/1+2]+[1/1+2+3]+[1/1+2+3+4]+...+[1/1+2+3+4+...+50
Tính tổng :\(S=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+....+\frac{1}{50}.\left(1+2+3+4+....+50\right)\)
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