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A = -1 + -2 + -3 + -4 + ... + -99 + -100
= - ( 1 + 2 +3 + ... + 100)
= - 5050
\(...\\ A=-\left(1+2+3+...+100\right)\\ A=-\left(\frac{\left(1+100\right).100}{2}\right)\\ A=-101.50=-5050\)
Chúc bạn học tốt!!!
Tui ra kết quả khác.
Tính nhanh:
\(\left(2^{100}+2^{101}+2^{102}\right):\left(2^{97}+2^{98}+2^{99}\right)\\ =2^3\left(2^{97}+2^{98}+2^{99}\right):\left(2^{97}+2^{98}+2^{99}\right)\\ =2^3=8\)
Giải:
\(\left(2^{100}+2^{101}+2^{102}\right):\left(2^{97}+2^{98}+2^{99}\right).\)
\(=\left(2^3.2^{97}+2^3.2^{98}+2^3.2^{99}\right):\left(2^{97}+2^{98}+2^{99}\right).\)
\(=2^3\left(2^{97}+2^{98}+2^{99}\right):\left(2^{97}+2^{98}+2^{99}\right).\)
\(=2^3\left[\left(2^{97}+2^{98}+2^{99}\right):\left(2^{97}+2^{98}+2^{99}\right)\right].\)
\(=2^3.1.\)
\(=2^3\left(=8\right).\)
~ Học tốt!!! ... ~ ^ _ ^
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bài 1 mifk viết sai nha.
bài 1: cho A=1+3+3\(^2\)+3\(^3\)+...+3\(^{10}\).Tìm số tự nhiên n biết 2 x A + 1 = 3\(^n\)
B1:
\(A=1+3+3^2+3^3+...+3^{10}\\ 3A=3+3^2+3^3+3^4+...+3^{11}\\ 3A-A=3^{11}-1\\ \Rightarrow A=\frac{3^{11}-1}{2}\)
mấy câu khác tương tự nha
1: =>7/3x=3+1/3-8-2/3=-5-1/3=-16/3
=>x=-16/3:7/3=-7/16
2: =>1/3|x-2|=4/5+3/7=28/35+15/35=43/35
=>|x-2|=129/35
=>x-2=129/35 hoặc x-2=-129/35
=>x=199/35 hoặc x=-59/35
Câu 1:
Đặt: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+....+\frac{1}{100^2}\)
\(=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+....+\frac{1}{100.100}\)
\(A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow A< \frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
Vậy:.............
Câu 2:
\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{98}+1\right)\left(\frac{1}{99}+1\right)\)
\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{98}+\frac{98}{98}\right)\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{99}{98}.\frac{100}{99}\)
\(=\frac{3.4.5....99.100}{2.3.4...98.99}\)
\(=\frac{100}{2}=50\)
Câu c :
2\(^{2x-1}\) - 2 = C
2\(^{2x-1}\)- 2 = 2\(^{101}\)- 2
2\(^{2x-1}\)= 2\(^{101}\)
2x - 1 = 101
2x = 101 + 1 = 102
x = \(\frac{102}{2}\)= 51
Vậy x = 51
a) \(C=2+2^2+2^3+..........+2^{99}+2^{100}\)
\(C=\left(2+2^2+2^3+2^4+2^5\right)+...............+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(C=1.\left(2+2^2+2^3+2^4+2^5\right)+...........+2^{96}.\left(2+2^2+2^3+2^4+2^5\right)\)
\(C=1.62+............+2^{96}.62\)
Mà 62 \(⋮\)31 \(\Rightarrow C⋮31\left(đpcm\right)\)
b) \(2C=2^2+2^3+2^4+2^5+2^6+...............+2^{100}+2^{101}\)
\(2C-C=\left(2^2+2^3+2^4...........+2^{100}+2^{101}\right)-\left(2+2^2+2^3..........2^{99}+2^{100}\right)\)
\(2C-C=2^2+2^3+2^4+...........+2^{100}+2^{101}-2-2^2-2^3-.........-2^{99}-2^{100}\)
\(C=2^{101}-2^{100}\)
c) 22x-1 - 2 = C
Bạn áp dụng phần b để làm
\(a,A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{2^{2018}}\)
\(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2016}}+\dfrac{1}{3^{2017}}\)
\(3A-A=1-\dfrac{1}{3^{2018}}\)
\(A=\dfrac{\left(1-\dfrac{1}{3^{2018}}\right)}{2}\)
\(b,B=1+5+5^2+5^3+...+5^{100}\)
\(5B=5+5^2+5^3+5^4+...+5^{100}+5^{101}\)
\(5B-B=1-5^{101}\)
\(B=\dfrac{\left(1-5^{101}\right)}{4}\)
Lời giải:
$B=2+2^2+2^3+...+2^{99}+2^{100}$
$2B=2^2+2^3+2^4+...+2^{100}+2^{101}$
$\Rightarrow 2B-B=(2^2+2^3+2^4+...+2^{100}+2^{101}) - (2+2^2+2^3+...+2^{99}+2^{100})$
$\Rightarrow B=2^{101}-2$
B=2+22+23+...+299+2100
2B=22+23+24+...+2100+2101
2B-B=2101-2
B=2101-2