Cho M bằng
M = 1 + 2 .(30 + 31 + 32 + ... + 398 + 399)
a)Thu gọn M
b) Tìm n để M = 9n
K
Khách
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Những câu hỏi liên quan
2 tháng 2 2021
a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)
\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)
\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)
\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)
AH
Akai Haruma
Giáo viên
5 tháng 2 2024
Bài 1:
a. $2^{29}< 5^{29}< 5^{39}$
$\Rightarrow A< B$
b.
$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$
$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$
$=(1+3)(3+3^3+3^5+...+3^{2009})$
$=4(3+3^3+3^5+...+3^{2009})\vdots 4$
Mặt khác:
$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$
$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$
$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$
AH
Akai Haruma
Giáo viên
5 tháng 2 2024
Bài 1:
c.
$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$
$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$
$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$
$\Rightarrow A=\frac{3^{101}+1}{4}$
NH
1
HQ
Hà Quang Minh
Giáo viên
1 tháng 8 2023
\(A=3^{100}-3^{99}+3^{98}-...-3+1\\ \Rightarrow\dfrac{1}{3}A=3^{99}-3^{98}+3^{97}-...-1+\dfrac{1}{3}\\ \Rightarrow\dfrac{4}{3}A=3^{100}+\dfrac{1}{3}\\ \Rightarrow A=\dfrac{3^{101}}{4}+\dfrac{1}{4}\)
2 tháng 10 2021
a: Tổng các số hạng là:
\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)
Ta có: A+1=2x
\(\Leftrightarrow2x=24311\)
hay \(x=\dfrac{24311}{2}\)
DT
2
19 tháng 10 2023
`#3107.101107`
\(A = 1 + 3 + 3^2 + 3^3 + ... + 3^{98} + 3^{99}\)
\(A = (1 + 3) + (3^2 + 3^3) + ... + (3^{98} + 3^{99})\)
\(A = (1 + 3) + 3^2(1 + 3) + ... + 3^{98}(1 + 3)\)
\(A = (1 + 3)(1 + 3^2 + ... + 3^{98})\)
\(A = 4(1 + 3^2 + ... + 3^{98})\)
Vì \(4(1 + 3^2 + ... + 3^{98}) \) \(\vdots\) \(4\)
`\Rightarrow A \vdots 4`
Vậy, `A \vdots 4` (đpcm).
19 tháng 10 2023
A = 1 + 3 + 32 + 33 + ... + 398 + 399
A = (1 + 3) + (32 + 33) + ... + (398 + 399)
A = 1. (1 + 3) + 32. (1 + 3) + ... + 398. (1 + 3)
A = 1.4 + 32.4 + ... + 398.4
A = 4. (1 + 32 + ... + 398)
⇒ A ⋮ 4
KJ
1
Q
13 tháng 12 2021
Tham khảo
Ta có: 3A = 3.(1+3+32+33+...+399+3100)(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+31013+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)(3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−13101−1
⇒⇒ A = 3101−123101−12
Vậy A = 3101−12
KJ
2
16 tháng 12 2021
\(A=1-3+3^2-3^3+3^4-...-3^{98}-3^{99}+3^{100}\\ 3A=3-3^2+3^3-3^4-...-3^{98}+3^{99}-3^{100}+3^{101}\\ 3A-A=3^{101}-1\\ \Rightarrow A=\dfrac{3^{101}-1}{2}\)
PL
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 12 2023
A = 1 - 3 + 32 - 33 + 34 - ... + 398 - 399 + 3100
3A = 3 - 32 + 33 - 34+ 35 - ... + 399 - 3100 + 3101
3A + A = 3 - 32+ 33-34+35 -...+399 - 3100 + 3101 + 1 - 3 +...-399+3100
4A = 3101 + 1
A = \(\dfrac{3^{101}+1}{4}\)
26 tháng 10 2023
a: \(A=3^{100}-3^{99}+3^{98}-...+3^2-3\)
=>\(3A=3^{101}-3^{100}+3^{99}-...+3^3-3^2\)
=>\(4A=3^{101}-3\)
=>\(A=\dfrac{3^{101}-3}{4}\)
b: \(B=\left(-2\right)^0+\left(-2\right)^1+...+\left(-2\right)^{2024}\)
=>\(B\cdot\left(-2\right)=\left(-2\right)^1+\left(-2\right)^2+...+\left(-2\right)^{2025}\)
=>\(-2B-B=\left(-2\right)^1+\left(-2\right)^2+...+\left(-2\right)^{2025}-\left(-2\right)^0-\left(-2\right)^1-...-\left(-2\right)^{2024}\)
=>\(-3B=-2^{2025}-1\)
=>\(B=\dfrac{2^{2025}+1}{3}\)
c: \(C=\left(-\dfrac{1}{5}\right)^0+\left(-\dfrac{1}{5}\right)^1+...+\left(-\dfrac{1}{5}\right)^{2023}\)
=>\(\left(-\dfrac{1}{5}\right)\cdot C=\left(-\dfrac{1}{5}\right)^1+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{2024}\)
=>\(\left(-\dfrac{6}{5}\right)\cdot C=\left(-\dfrac{1}{5}\right)^{2024}-\left(-\dfrac{1}{5}\right)^0\)
=>\(C\cdot\dfrac{-6}{5}=\dfrac{1}{5^{2024}}-1=\dfrac{1-5^{2024}}{5^{2024}}\)
=>\(C\cdot\dfrac{6}{5}=\dfrac{5^{2024}-1}{5^{2024}}\)
=>\(C=\dfrac{5^{2024}-1}{5^{2024}}:\dfrac{6}{5}=\dfrac{5^{2024}-1}{6\cdot5^{2023}}\)