1/Tính A
A = 1/2 + 3/2 + (3/2)2 + (3/2)3 + .....+ (3/2)2016
2/ Cho : S = 1/4 +2/42 + 3/43 +....+ 2014/42014
Chứng minh rằng : S < 1/2
Bài 1. So sánh: \(2^{49}\) và \(5^{21}\)
Bài 2. a, Chứng minh rằng S = 1 + 3 + 32 + 33 + ... + 399 chia hết cho 40.
b, Cho S = 1 + 4 + 42 + 43 + ... + 462. Chứng minh rằng S chia hết cho 21.
Giúp mk với
Bài 1:
\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)
Bài 2:
\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)
Bài 1 :
\(2^{49}=\left(2^7\right)^7=128^7\)
\(5^{21}=\left(5^3\right)^7=125^7\)
mà \(125^7< 128^7\)
\(\Rightarrow2^{49}>5^{21}\)
Bài 2 :
a) \(S=1+3+3^2+3^3+...3^{99}\)
\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)
\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)
\(\Rightarrow dpcm\)
b) \(S=1+4+4^2+4^3+...4^{62}\)
\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)
\(\Rightarrow S=21+4^3.21+...4^{60}.21\)
\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)
\(\Rightarrow dpcm\)
1.Tính: A=3/5+3/5^4+3/5^7+...+3/5^100
2.Chứng minh rằng: 1/3+2/3^2+3/3^3+4/3^4+5/3^5+...+100/3^100<3/4
3. Tính: S=a+a^2+a^3+a^4+...a^2022
B=a-a^2+a^3-a^4+...-a^2022
giúp mk vs ak :3
Bài 3:
a: a*S=a^2+a^3+...+a^2023
=>(a-1)*S=a^2023-a
=>\(S=\dfrac{a^{2023}-a}{a-1}\)
b: a*B=a^2-a^3+...-a^2023
=>(a+1)B=a-a^2023
=>\(B=\dfrac{a-a^{2023}}{a+1}\)
Cho S = 1/4 + 2/4^2 + 3/4^3 + . . . + 2014/4^2014
Chứng minh rằng : S < 1/2
Đang cần gấp :33
a)1^2/1×2×2^2/2×3/3^2/3×4×...×999^2/999×1000
b)A=1/101+1/102+1/103+...+1/150. CM: 1/3 <A <1/2
c) So sánh: A= 2011+2012/2012+2013 và B=2011/2012+2012/2013
d) So sánh: S= 1/11+1/12+1/13+...+1/20 và 1/2
e) CM: 7/12<1/41+1/42+1/43+...+1/80 <1
f) So sánh: A= 2^2014+1/2^2014 và B= 2^2014+2/2^2014+1
g) Rút gọn: B= (1-1/2)×(1-1/3)×(1-1/4)×...×(1-1/20)
h) (1+1/2)×(1+1/3)+(1+1/4)×...×(1+1/99)
Các bạn chỉ cần làm những câu hỏi các bạn biết thôi nha. Mình đang cần gấp.
g: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{19}{20}=\dfrac{1}{20}\)
h: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot..\cdot\dfrac{100}{99}=\dfrac{100}{2}=50\)
f: \(A=1+\dfrac{1}{2^{2014}}\)
\(B=\dfrac{2^{2014}+1+1}{2^{2014}+1}=1+\dfrac{1}{2^{2014}+1}\)
mà \(2^{2014}< 2^{2014}+1\)
nên A>B
2. a)S=1-2+2^2-2^3+...........+2^2014 tính S.
b) So sánh: A=2^2013+3/2^2014+3 và B=2^2014+3/2^2015+3.
c) tìm các số tự nhiên a,b :a/3+b/4=a+b/3+4.
3. tìm các số tự nhiên x,y biết: (2^x+1) (2^x+2) (2^x+3) (2^x+4)-5^y=11879.
Bài 1: Cho A= 2 + 2 ^ 2 + 2 ^ 3 +.......+2^ 60 . Chứng tỏ rằng: 4 chia hết cho 3,5,7. Bài 2: Cho S= 1 + 5 ^ 2 + 5 ^ 4 + 5 ^ 6 +***+5^ 2020 . Chứng minh rằng S chia hết cho 313 Bài 3: Tính A= 5 + 5 ^ 2 + 5 ^ 3 +...+5^ 12
Bài 3:
\(A=5+5^2+..+5^{12}\)
\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)
\(5A=5^2+5^3+...+5^{13}\)
\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)
\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)
\(4A=5^{13}-5\)
\(A=\dfrac{5^{13}-5}{4}\)
Cho tổng gồm 2014 số hạng: \(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2014}{4^{2014}}\). Chứng minh rằng \(S< \dfrac{1}{2}\)
4S=1+24+342+....+2014420134S=1+24+342+....+201442013
4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)
3S=1+(24−14)+(342−242)+......+(201442013−201342013)−2014420143S=1+(24−14)+(342−242)+......+(201442013−201342013)−201442014
3S=1+14+142+143+.....+142013−2014420143S=1+14+142+143+.....+142013−201442014
đặt A=1+14+142+143+....+142023A=1+14+142+143+....+142023
4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)
3A=4−1420233A=4−142023
A=43−13.42023A=43−13.42023
⇒3S=43−13.42023−201442024⇒3S=43−13.42023−201442024
⇒S=49−19.42023−20143.42024⇒S=49−19.42023−20143.42024
do 49<48=1249<48=12
⇒S=49−19.42023−20143.42024<48=12(đpcm)
S=(1+1/2)+(1+1/22)+(1+3/23)+...+(1+2014/22014)
Chứng minh rằng S<2016
tính tổng
1. A =1/1^2+1/2^2+1/3^2+1/4^2+...+1/50^2
chứng minh rằng A <2
2. S=3+3/2+3/2^2+3/2^4+...+3/2^9
A=\(\frac{1}{1^2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)
A=1+\(\frac{1}{2^2}\)\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)
A<1+\(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+...+\(\frac{1}{49\cdot50}\)
A<1+1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
A<2-\(\frac{1}{50}\)<2
=>A<1(câu 1)