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NA
1
14 tháng 6 2023
\(B=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{50\cdot51}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{51}{2}\)
\(=\dfrac{50\cdot\dfrac{\left(51+2\right)}{2}}{2}=50\cdot\dfrac{53}{4}=662.5\)
NN
0
I
0
TB
3
S
23 tháng 6 2019
\(P=1+\frac{1}{1+2}+\frac{1}{1+2+3}+........+\frac{1}{1+2+3+.......+50}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+......+\frac{1}{\frac{50.51}{2}}=1+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{50.51}=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{50.51}\right)\) \(Taco:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\)
\(\Rightarrow P=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.......+\frac{1}{50}-\frac{1}{51}\right)=1+2\left(\frac{1}{2}-\frac{1}{51}\right)=1+1-\frac{2}{51}=2-\frac{2}{51}=\frac{100}{51}\)
VD
1
TV
5
3 tháng 1 2016
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 \(S=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)
AH
Akai Haruma
Giáo viên
23 tháng 2 2021
Lời giải:
\(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}>\frac{1}{4}+\frac{1}{3.4}+\frac{1}{4.5}....+\frac{1}{50.51}\)
Mà:
\(\frac{1}{4}+\frac{1}{3.4}+\frac{1}{4.5}....+\frac{1}{50.51}=\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{50}-\frac{1}{51}=\frac{7}{12}-\frac{1}{51}>\frac{1}{2}\)
Do đó: $B>\frac{1}{2}$
Ta có đpcm.
23 tháng 2 2021
Lời giải:
B=122+132+...+1502>14+13.4+14.5....+150.51B=122+132+...+1502>14+13.4+14.5....+150.51
Mà:
14+13.4+14.5....+150.51=14+13−14+....+150−151=712−151>1214+13.4+14.5....+150.51=14+13−14+....+150−151=712−151>12
Do đó: B>12B>12
Ta có đpcm.
MH
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 7
A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}\)+ \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)
A = 1
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{50^2}\right)=\frac{3}{2^2}.\frac{8}{3^2}...\frac{2499}{50^2}=\frac{1.3.2.4...49.51}{2.2.3.3...50.50}\)
\(=\frac{\left(1.2...49\right)\left(3.4...51\right)}{\left(2.3...50\right)\left(2.3....50\right)}=\frac{1.51}{50.2}=\frac{51}{100}\)