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Lời giải:
a) \(A=1+3+3^2+3^3+...+3^{100}\)
\(\Rightarrow 3A=3+3^2+3^3+...+3^{101}\)
Trừ theo vế:
\(\Rightarrow 3A-A=(3+3^2+3^3+..+3^{101})-(1+3+3^2+...+3^{100})\)
\(2A=3^{101}-1\Rightarrow A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow 2B=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
Cộng theo vế:
\(\Rightarrow B+2B=2^{201}-2\)
\(\Rightarrow B=\frac{2^{101}-2}{3}\)
c) Ta có:
\(C=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
\(\Rightarrow 3C=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)
Cộng theo vế:
\(C+3C=(3^{100}-3^{99}+3^{98}-....+3^2-3+1)+(3^{101}-3^{100}+3^{99}-....+3^3-3^2+3)\)
\(4C=3^{101}+1\Rightarrow C=\frac{3^{101}+1}{4}\)
a: \(3A=3+3^2+...+3^{101}\)
\(\Leftrightarrow2A=3^{101}-1\)
hay \(A=\dfrac{3^{101}-1}{2}\)
b: \(2B=2^{101}-2^{100}+...+2^3-2^2\)
\(\Leftrightarrow3B=2^{101}-2\)
hay \(B=\dfrac{2^{101}-2}{3}\)
c: \(3C=3^{101}-3^{100}+....+3^3-3^2+3\)
=>\(4C=3^{101}+1\)
hay \(C=\dfrac{3^{101}+1}{4}\)
Ta có :
\(A=\frac{1}{3}+\frac{2}{3^2}+......+\frac{100}{3^{100}}\) \(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+.....+\frac{100}{3^{99}}\)
\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)= 2A
Đặt \(B=1+\frac{1}{3}+...+\frac{1}{3^{99}}\) \(\Rightarrow3B=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\)
\(\Rightarrow3B-B=3-\frac{1}{3^{99}}=2B\) \(\Rightarrow B=\frac{3}{2}-\frac{1}{3^{99}.2}\)
\(\Rightarrow2A=\frac{3}{2}-\frac{1}{3^{99}.2}-\frac{100}{3^{100}}\)\(\Rightarrow A=\frac{3}{4}-\frac{1}{3^{99}.4}-\frac{100}{3^{100}}< \frac{3}{4}\Rightarrow\left(đpcm\right)\)
Ta có :
\(C=1+3+3^2+....+3^{100}\) \(\Rightarrow C-1=3+3^2+....+3^{100}\)
\(\Rightarrow3\left(C-1\right)=3^2+3^3+.....+3^{101}\)\(\Rightarrow3C-3-\left(C-1\right)=3^{101}-3\)
\(\Rightarrow2C-2=3^{101}-3\Rightarrow2C=3^{101}-1\)\(\Rightarrow C=\frac{3^{101}-1}{2}\)
Ta có :
\(D=2^{100}-2^{99}+2^{98}-.....-2\) \(\Rightarrow2D=2^{101}-2^{100}+2^{99}-.....-2^2\)
\(\Rightarrow2D+D=2^{101}-2=3D\) \(\Rightarrow D=\frac{2^{101}-2}{3}\)
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(2A=1+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)-\frac{100}{3^{100}}\)
Ta thấy biểu thức trong dấu ngoặc nhỏ hơn 1/2 ( tự chứng minh ) nên 2A < 1 + 1/2
\(\Rightarrow A< \frac{3}{4}\)
\(C=1+3+3^2+3^3+...+3^{100}\)
\(3C=3+3^2+3^3+3^4+...+3^{101}\)
\(3C-C=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)
\(2C=3^{101}-1\)
\(C=\frac{3^{101}-1}{2}\)
1) \(+2x+3y⋮17\)
\(\Rightarrow26x+39y⋮17\)
\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)
Mà \(17x+34y⋮17\)
\(\Rightarrow9x+5y⋮17\)
\(+9x+5y⋮17\)
\(\Rightarrow36x+20y⋮17\)
\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)
Mà \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
\(A=1+3+3^2+3^3+...+3^{99}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A-A=2A=\left(3+3^2+3^3+...+3^{100}\right)-\left(\text{}\text{}\text{}1+3^2+3^3+...+3^{99}\right)\)
\(\Rightarrow2A=3^{100}-1\Rightarrow A=\frac{3^{100}-1}{2}\)
C=1+3+32+.............+3100
C=\(\frac{3C-C}{2}\)
3C=3+32+33+.............+399+3100+3101
C=1+3+32+..................+399+3100
3C-C=(3+32+33+.............+399+3100+3101)-(1+3+32+..................+399+3100)
Triệt tiêu các số hạng co giá trị tuyệt đối bằng nhau, ta được:
2C=-1+3100
\(\Rightarrow C=\frac{3^{100}-1}{2}\)
D=\(\frac{2D+D}{3}\)
2D=2101-2100+299-298+..............+23-22
D=2100-299+298-297+............+22-2
2D+D=2101-2100+299-298+..............+23-22+2100-299+298-297+............+22-2
Triệt tiêu các số hạng có giá trị tuyệt đối bằng nhau, ta được:
3D=2101-2
\(\Rightarrow D=\frac{2^{101}-2}{3}\)
B=\(\frac{3}{1\times4}+\frac{5}{4\times9}+\frac{7}{9\times16}+.........+\frac{19}{81\times100}\)
Quan sát biểu thức, ta có nhận xét:
4-1=3;
9-4=5;
16-9=7;
.......;100-81=19
=> Hiệu hai số ở mẫu bằng giá trị ở tử
\(\Rightarrow B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+.......+\frac{1}{81}-\frac{1}{100}\)
\(\Rightarrow B=1-\frac{1}{100}\)
\(B=\frac{99}{100}< \frac{100}{100}\)
Vậy B<1
A = 2100 - 299 + 298 - 297 + ... + 22 - 2
= ( 2100 + 298 + ... + 22 ) - ( 299 + 297 + ... + 2 )
= ( 2100 + 298 + ... + 22 ) - 2( 299 + 297 + ... + 2 ) + ( 299 + 297 + ... + 2 )
= 299 + 297 + ... + 2
=> 4A = 2103 + 299 + ... + 23
=> 3A = 2103 - 2
=> A = \(\frac{2^{103}-2}{3}\)
help me
a) Ta có: \(A=1^3+2^3+3^3+...+100^3\)
\(=\left(1-1\right)\cdot1\cdot\left(1+1\right)+1+\left(2-1\right)\cdot2\cdot\left(2+1\right)+2+...+\left(100-1\right)\cdot100\cdot\left(100+1\right)+100\)
\(=1+2+1\cdot2\cdot3+...+99\cdot100\cdot101\)
\(=5050+25497450\)
\(=25502500\)