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Bài 1
a) S = 1 + 2 + 2² + 2³ + ... + 2²⁰²³
2S = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²⁴
S = 2S - S = (2 + 2² + 2³ + ... + 2²⁰²⁴) - (1 + 2 + 2² + 2³)
= 2²⁰²⁴ - 1
b) B = 2²⁰²⁴
B - 1 = 2²⁰²⁴ - 1 = S
B = S + 1
Vậy B > S
a,
\(S=1+2+2^2+...+2^{2023}\)
\(2S=2+2^2+2^3+...+2^{2024}\)
\(\Rightarrow S=2^{2024}-1\)
b.
Do \(2^{2024}-1< 2^{2024}\)
\(\Rightarrow S< B\)
2.
\(H=3+3^2+...+3^{2022}\)
\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)
\(\Rightarrow3H-H=3^{2023}-3\)
\(\Rightarrow2H=3^{2023}-3\)
\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)
A = 1 + 3 + 3² + ... + 3²⁰²³
⇒ 3A = 3 + 3² + 3³ + ... + 3²⁰²³ + 3²⁰²⁴
⇒ 2A = 3A - A
= (3 + 3² + 3³ + ... + 3²⁰²³ + 3²⁰²⁴) - (1 + 3 + 3² + ... + 3²⁰²³)
= 3²⁰²⁴ - 1
⇒ A = (3²⁰²⁴ - 1) : 2
⇒ A < B
A=1+3+32+33+34+........+32022+32023
3A=3+32+33+............+32023+32024
3A-A=(3+32+33+..........+32023+32024
= 1/2.2 + 1/3.3 + ... + 1/2018.2018
= ( 1/2 - 1/2) + (1/3 - 1/3) + ... + ( 1/2018 - 1/2018 )
= 0+0+0+0+...+0
=0
75% = 7,5
7,5 > 0 ==>
A<B
B = 75% => B = 3/4
Ta có :\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}=1-\frac{1}{2018}\)
Vì \(\frac{1}{2018}< \frac{1}{4}\Rightarrow1-\frac{1}{2018}>1-\frac{1}{4}\Rightarrow A>\frac{3}{4}\)=> A > B
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}\)
\(B=75\%=\frac{3}{4}\)
Ta có:\(A=.......\)
\(=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\right)< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2018}=\frac{3}{4}-\frac{1}{2018}< \frac{3}{4}\)
\(\Rightarrow A< B\)
A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰
⇒ 2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹
⇒ A = 2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰)
= 2²⁰¹¹ - 2⁰
= 2²⁰¹¹ - 1
= B
Vậy A = B
Sửa đề: so sánh với 1/2
1/3^2<1/2*3
1/4^2<1/3*4
...
1/80^2<1/79*80
=>1/3^2+1/4^2+...+1/80^2<1/2-1/3+1/3-1/4+...+1/79-1/80=39/80<1/2