cho A=1/12+ 1/22+1/33+...+1/1010
tính A?
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Ta thấy:
1/7 + 1/14 + 1/28 = (4+2+1)/28 = 1/4
1/8 + 1/12 + 1/24 = (3+2+1)/24 = 1/4
1/11 + 1/22 + 1/33 = (6+3+2)/66 = 1/6
1/9 + 1/18 = (2+1)/18 = 1/6
Mà 1/4 + 1/4 = 1/2 và 1/6 + 1/6 = 1/3
Nên:
1/10 + 1/13 + 1/15 + 1/2 + 1/3 =
1/10 + 1/13 + 1/15 + 5/10 + 5/15 =
6/10 + 6/15 + 1/13 =
3/5 + 2/5 + 1/13 =
1 + 1/13 = 14/13
TT ik nha bạn ))))
a)Tính bằng cách thuận tiện nhất:
1/7+1/8+1/9+1/10+1/11+1/12+1/13+1/14+1/15+1/18+1/22+1/24+1/28+1/33
Giải
Ta thấy:
1/7 + 1/14 + 1/28 = (4+2+1)/28 = 1/4
1/8 + 1/12 + 1/24 = (3+2+1)/24 = 1/4
1/11 + 1/22 + 1/33 = (6+3+2)/66 = 1/6
1/9 + 1/18 = (2+1)/18 = 1/6
Mà 1/4 + 1/4 = 1/2 và 1/6 + 1/6 = 1/3
Nên:
1/10 + 1/13 + 1/15 + 1/2 + 1/3 =
1/10 + 1/13 + 1/15 + 5/10 + 5/15 =
6/10 + 6/15 + 1/13 =
3/5 + 2/5 + 1/13 =
1 + 1/13 = 14/13
1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{97}\right)\)
\(=30\left(1+2^4+...+2^{96}\right)⋮30\)
2:
\(B=3+3^2+3^3+...+3^{2022}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)
\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)
1/3 + 1/2 + 1/5 + 12/15 + 22/33 + 16/32
=1/3 + 1/2 + 1/5 + 4/5 + 2/3 + 1/2
=(1/3 + 2/3) + (1/5 + 4/5) + (1/2 + 1/2)
= 1 + 1 + 1
=3
Chúc bạn học tốt !
Phần ở dưới mới là câu hỏi đúng phần câu hỏi ở trên vẫn còn thiếu nha !
a) P = 1 + 3 + 3² + ... + 3¹⁰¹
= (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3⁹⁹ + 3¹⁰⁰ + 3¹⁰¹)
= 13 + 3³.(1 + 3 + 3²) + ... + 3⁹⁹.(1 + 3 + 3²)
= 13 + 3³.13 + ... + 3⁹⁹.13
= 13.(1 + 3³ + ... + 3⁹⁹) ⋮ 13
Vậy P ⋮ 13
b) B = 1 + 2² + 2⁴ + ... + 2²⁰²⁰
= (1 + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰)
= 21 + 2⁶.(1 + 2² + 2⁴) + ... + 2²⁰¹⁶.(1 + 2² + 2⁴)
= 21 + 2⁶.21 + ... + 2²⁰¹⁶.21
= 21.(1 + 2⁶ + ... + 2²⁰¹⁶) ⋮ 21
Vậy B ⋮ 21
c) A = 2 + 2² + 2³ + ... + 2²⁰
= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁷ + 2¹⁸ + 2¹⁹ + 2²⁰)
= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2¹⁶.(2 + 2² + 2³ + 2⁴)
= 30 + 2⁴.30 + ... + 2¹⁶.30
= 30.(1 + 2⁴ + ... + 2¹⁶)
= 5.6.(1 + 2⁴ + ... + 2¹⁶) ⋮ 5
Vậy A ⋮ 5
d) A = 1 + 4 + 4² + ... + 4⁹⁸
= (1 + 4 + 4²) + (4³ + 4⁴ + 4⁵) + ... + (4⁹⁷ + 4⁹⁸ + 4⁹⁹)
= 21 + 4³.(1 + 4 + 4²) + ... + 4⁹⁷.(1 + 4 + 4²)
= 21 + 4³.21 + ... + 4⁹⁷.21
= 21.(1 + 4³ + ... + 4⁹⁷) ⋮ 21
Vậy A ⋮ 21
e) A = 11⁹ + 11⁸ + 11⁷ + ... + 11 + 1
= (11⁹ + 11⁸ + 11⁷ + 11⁶ + 11⁵) + (11⁴ + 11³ + 11² + 11 + 1)
= 11⁵.(11⁴ + 11³ + 11² + 11 + 1) + 16105
= 11⁵.16105 + 16105
= 16105.(11⁵ + 1)
= 5.3221.(11⁵ + 1) ⋮ 5
Vậy A ⋮ 5
1/3+1/4+1/5+/12/15+22/33+48/64
=1/3+1/4+1/5+4/5+2/3+3/4
=3
dung 100% luon
Ai k cho minh 3 k thi minh se k cho nguoi ay 5 k
a: Tổng các số hạng là:
\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)
Ta có: A+1=2x
\(\Leftrightarrow2x=24311\)
hay \(x=\dfrac{24311}{2}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{4}\)
=(1/3+2/3)+(1/4+1/4)+(1/5+4/5)
=1+1+1/2
=5/2
a/ 1+4+7+10+....+52+55+58-410=[(58-1):3+1]x(58+1)-410=20x59-410=770