a)Tính tổng:S=1/1.2.3+1/2.3.4+...+1/98.99.100
b)Chứng minh:A=1/2.(1/6+1/24+1/60+...+1/9240)>57/462
a) Tính tổng \(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
b) Chứng minh: \(A=\frac{1}{2}+\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{9240}\right)>\frac{57}{462}\)
Làm lại câu a
\(2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(2S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(2S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(2S=1-\frac{1}{100}\)suy ra \(2S=\frac{99}{100}\)
\(S=\frac{99}{100}:2\)suy ra \(S=\frac{99}{200}\)
a, 2S=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(2S=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\)
\(2S=1-\frac{1}{100}\)suy ra \(2S=\frac{99}{100}\)
\(S=\frac{99}{100}:2=\frac{99}{200}\)
Ta có 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/98.99.100
= 1/2 ( 1 / 1.2 - 1 / 2.3 + 1 / 2.3 - 1 / 3.4 + ...................+ 1 / 97.98 - 1 / 98.99 + 1 / 98.99 - 1 / 99.100)
= 1 / 2 ( 1 / 1.2 - 1 / 99.100 )
= 4949 / 19800
a) Tính tổng \(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
b) Chứng minh: \(A=\frac{1}{2}+\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{9240}\right)>\frac{57}{462}\)
4 lke
Ta xét: \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3};\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4};...;\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)
Tổng quát : \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\). Do đó:
\(2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)-...-\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}\)
Vậy \(S=\frac{4949}{9900}\)
b, Ta có : \(\frac{1}{2}>\frac{57}{462}\)mà \(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{9240}>0\)
nên A = \(\frac{1}{2}+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{9240}\right)>\frac{57}{462}+0=\frac{57}{462}\)
Bạn sai Câu a.\(2S=\frac{4949}{9900}\)Vậy \(S=\frac{4949}{9900}:2=\frac{4949}{9900}\cdot\frac{1}{2}=\frac{4949}{19800}\)
Chứng minh rằng A=1/2(1/6+1/24+1/60+....+1/9240)>57/462
chứng minh rằng : A=1/2.(1/6+1/24+1/60+...+1/9240)>57/462
Chứng minh rằng : A=1/2.(1/6+1/24+1/60+...+1/9240)>57/462
A= 1/2 ( 1/6+1/24+1/60+....+ 1/9240) > 57/462
cmr: A= 1/2.(1/6+1/24+1/60+...+1/9240)>57/462
So sánh A=1/2(1/6+1/24+1/60+...........+1/9240) và 57/462
CHỨNG MINH : A =1/2*(1/6+1/24+1/60+...+1/9240) > 57/462
giúp mik giải bài này nhé
Xét riêng cái biểu thức trong ngoặc nhé, đặt là B :
B = 1/6+1/24+1/60+...+1/9240
B = 1/1x2x3+1/2x3x4+1/3x4x5+...+1/20x21x22
B = (1/1x2 - 1/2x3) : 2 + (1/2x3-1/3x4) : 2 + (1/3x4 - 1/4x5) : 2 +...+ (1/20x21 - 1/21x22) : 2
B = (1/1x2-1/2x3+1/2x3-1/3x4+1/3x4-...-1/21x22) : 2
B = (1/2 - 1/462) : 2
B = 115/462
=> A = 1/2 x 115/462 > 1/2 x 114/462 = 57/462 (đpcm)