Tính tổng: S=1/3+1/32+1/33+...+1/38+1/39
So sánh tổng S= 1/31+1/32+1/33+1/34+1/35+1/36+1/37+1/38+1/39+1/40 với 1/4
Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)
Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)
\(\dfrac{1}{32}>\dfrac{1}{40}\)
\(\dfrac{1}{33}>\dfrac{1}{40}\)
\(\dfrac{1}{34}>\dfrac{1}{40}\)
\(\dfrac{1}{35}>\dfrac{1}{40}\)
\(\dfrac{1}{36}>\dfrac{1}{40}\)
\(\dfrac{1}{37}>\dfrac{1}{40}\)
\(\dfrac{1}{38}>\dfrac{1}{40}\)
\(\dfrac{1}{39}>\dfrac{1}{40}\)
\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)
Vậy \(S>\dfrac{1}{4}\)
Cho S = 1 + 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39. Chứng tỏ rằng S chia hết cho 4.
\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)
Cho S = 1 + 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39.Chứng tỏ rằng S chia hết cho 13.
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
Cho S = 1 + 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39. Chứng tỏ rằng S chia hết cho 4.
\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)
\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)
\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)
\(S=4\left(3^2+3^4+3^6+3^8\right)\)
\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)
Cho S = 1+3+32+33+34+35+36+37+38+39.Chứng tỏ rằng S chia hết cho 4
Giup mik vs
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4
S = 1 + 3 + 32 + 33 +... + 32014 .Tính tổng
S = 1 + 3 + 32 + 33 +... + 32014
3S = 3 + 32 + 33 + 34 + ... + 32015
3S - S = ( 3 + 32 + 33 + 34 + ... + 32015) - (1 + 3 + 32 + 33 +... + 32014)
2S = 32015 - 1
S = \(\dfrac{3^{2015}-1}{2}\)
S=1/2+1/3+...+1/38+1/39,P=1/39+2/38+...+38/2+39/1 tính S/P
bài 1:cho S = 1+2+22+23+...+22023
a. tính tổng
b.cho B = 22024 so sánh S và B
bài 2: tính tổng H=3+32+33+...+32022
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.
\(H=3+3^2+...+3^{2022}\)
\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)
\(\Rightarrow3H-H=3^{2023}-3\)
\(\Rightarrow2H=3^{2023}-3\)
\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)
Bài 2
H = 3 + 3² + 3³ + ... + 3²⁰²²
⇒ 3H = 3² + 3³ + 3⁴ + ... + 3²⁰²³
⇒2H = 3H - H
= (3² + 3³ + 3⁴ + ... + 3²⁰²³) - (3 + 3² + 3³ + ... + 3²⁰²²)
= 3²⁰²³ - 3
⇒ H = (3²⁰²³ - 3) : 2
tính tổng S = 1 + 31 + 32 + 33 + ..... + 3101
`#3107.101107`
\(S=1+3^1+3^2+3^3+...+3^{101}\)
\(3S=3+3^2+3^3+...+3^{102}\)
\(3S-S=\left(3+3^2+3^3+...+3^{102}\right)-\left(1+3+3^2+...+3^{101}\right)\)
\(2S=3+3^2+3^3+3^{102}-1-3-3^2-...-3^{101}\)
\(2S=3^{102}-1\)
\(S=\dfrac{3^{102}-1}{2}\)
Vậy, \(S=\dfrac{3^{102}-1}{2}.\)
3s=3+3^2+3^3+....+3^102
3s-s=2s
2s=3^102-1
s=3^102-1 trên2
tính:
1+3+32+33+...+39
S = 1 + 3 + 32 + 33 +...+39
3.S = 3 + 32 + 33 +....+39+310
3S-S = 310 - 1
2S = 310 - 1
S = \(\dfrac{3^{10}-1}{2}\)