\(B=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{50\cdot51}{2}\)

\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{51}{2}\)

\(=\dfrac{50\cdot\dfrac{\left(51+2\right)}{2}}{2}=50\cdot\dfrac{53}{4}=662.5\)

\(E=-\dfrac{1}{3}\cdot\left(1+2+3\right)-\dfrac{1}{4}\left(1+2+3+4\right)-...-\dfrac{1}{50}\left(1+2+3+...+50\right)\)

\(=\dfrac{-1}{3}\cdot\dfrac{3\cdot4}{2}-\dfrac{1}{4}\cdot\dfrac{4\cdot5}{2}-...-\dfrac{1}{50}\cdot\dfrac{50\cdot51}{2}\)

\(=\dfrac{-4}{2}-\dfrac{5}{2}-...-\dfrac{51}{2}\)

\(=\dfrac{-\left(4+5+...+51\right)}{2}\)

\(=\dfrac{-\left(51+4\right)\cdot\dfrac{48}{2}}{2}=-\dfrac{1320}{2}=-660\)

Đa thức chia x-1 có ngiệm là 1 nên:

Thay x=1 vào đa thức chia ta có:

1³⁰+1⁴-1¹⁹⁷⁵+1

=1+1-1+1

=2

Vậy số dư khi chia khi chia x³⁰+x⁴-x¹⁹⁷⁵+1 cho x-1 là 2

mk nghĩ là -1

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 \(S=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)

cách tính như thế nào bạn?????

Ahhh...trình bày dài lắm,cách làm thôi nhé

bạn vào nick ''nguyen thi thanh loan'' nhé

Là sao vậy

=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

\(P=1+\frac{1}{1+2}+\frac{1}{1+2+3}+........+\frac{1}{1+2+3+.......+50}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+......+\frac{1}{\frac{50.51}{2}}=1+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{50.51}=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{50.51}\right)\) \(Taco:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\)

\(\Rightarrow P=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.......+\frac{1}{50}-\frac{1}{51}\right)=1+2\left(\frac{1}{2}-\frac{1}{51}\right)=1+1-\frac{2}{51}=2-\frac{2}{51}=\frac{100}{51}\)

Bằng \(\frac{100}{51}\)

#)Giải :

Đặt \(A=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+50}\)

\(\Rightarrow A=\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{50.50:2}\)

\(\Rightarrow A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{100.101}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{51}\)

\(\Rightarrow\frac{1}{2}A=\frac{49}{102}\)

\(\Rightarrow A=\frac{49}{102}.\frac{1}{2}\)

\(\Rightarrow A=\frac{49}{204}\)