A=1/1^2+1/2^2+1/3^2+1/4^2+...+1/50^2
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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
DV
1
N
23 tháng 7 2015
a)\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(A=1-\frac{1}{2^{50}}<1\)
Vậy \(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}<1\)
b)\(B=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3B-B=2B=1-\frac{1}{3^{100}}\)
\(B=\frac{1-\frac{1}{3^{100}}}{2}\)
Vì \(1-\frac{1}{3^{100}}<1\)nên\(\frac{1-\frac{1}{3^{100}}}{2}<\frac{1}{2}\)
Vậy \(B=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}<\frac{1}{2}\)
c) \(C=\frac{1}{4^1}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}\)
\(4C=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{998}}+\frac{1}{4^{999}}\)
\(4C-C=3C=1-\frac{1}{4^{1000}}\)
\(C=\frac{1-\frac{1}{4^{1000}}}{3}\)
Vì \(1-\frac{1}{4^{1000}}<1\)nên\(\frac{1-\frac{1}{4^{1000}}}{3}<\frac{1}{3}\)
Vậy \(C=\frac{1}{4^1}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}<\frac{1}{3}\)
9 tháng 10 2016
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
V
2
6 tháng 8 2018
A = 1 × 2 × 3 + 2 × 3 × 4 + .....+ 48 × 49 × 50
ta có 4 x A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x (5 -1) + .....+ 48 × 49 × 50 x (51 - 47)
= 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + ... + 48 x 49 x 50 x 51 - 47 x 48 x 49 x 50
= 48 x 49 x 50 x 51
suy ra A = (48 x 49 x 50 x 51) : 4
= 12 x 49 x 50 x 51
nhớ k cho mik nha rùi mik lm nốt cho
LK
LÃ KHÔI NGUYÊN
VIP
12 tháng 10 2024
A = 1 × 2 × 3 + 2 × 3 × 4 + .....+ 48 × 49 × 50
ta có 4 x A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x (5 -1) + .....+ 48 × 49 × 50 x (51 - 47)
= 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + ... + 48 x 49 x 50 x 51 - 47 x 48 x 49 x 50
= 48 x 49 x 50 x 51
suy ra A = (48 x 49 x 50 x 51) : 4
= 12 x 49 x 50 x 51
TM
1
8 tháng 12 2023
\(E=-\dfrac{1}{3}\cdot\left(1+2+3\right)-\dfrac{1}{4}\left(1+2+3+4\right)-...-\dfrac{1}{50}\left(1+2+3+...+50\right)\)
\(=\dfrac{-1}{3}\cdot\dfrac{3\cdot4}{2}-\dfrac{1}{4}\cdot\dfrac{4\cdot5}{2}-...-\dfrac{1}{50}\cdot\dfrac{50\cdot51}{2}\)
\(=\dfrac{-4}{2}-\dfrac{5}{2}-...-\dfrac{51}{2}\)
\(=\dfrac{-\left(4+5+...+51\right)}{2}\)
\(=\dfrac{-\left(51+4\right)\cdot\dfrac{48}{2}}{2}=-\dfrac{1320}{2}=-660\)
MT
2
IW
11 tháng 5 2016
\(\Rightarrow A<1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{49.50}\)
\(\Rightarrow A<1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow A<1+\left(1-\frac{1}{50}\right)\)
\(\Rightarrow A<1+\frac{49}{50}\)
\(\Rightarrow A<\frac{99}{50}\)
Vì \(\frac{99}{50}<2=\frac{100}{50}\Rightarrow A<2\) ĐPCM
HP
11 tháng 5 2016
Ta có:
\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{50^2}<\frac{1}{49.50}\)
Do đó \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}<1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(\Rightarrow A<1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}<2\)
=>A<2(đpcm)
ND
0