\(1^2+2^2+3^2+...+99^2+100^2\). 

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8 tháng 2 2024

=1+ 2 (1+1)+ (2+1 )3+...+(99+1)100

=1+2+1.2+2.3+3+...+99.100+100

=(1+2+3+...+100)+(1.2+2.3+...+99.100)

=5050+(1.2+2.3+...+99.100)

đặt A=1.2+2.3+...+99.100

=>3A=1.2.3+2.3.3+...+99.100.3

=1.2.3+2.3(4-1)+...+99.100(101-98)

=1.2.3-1.2.3+2.3.4-2.3.4+...+98.99.100-98.99.100+99.100.101

=999900

=>A=333300

=>M=333300+5050=338350

15 tháng 7 2019

\(a,A=1^2+3^2+5^2+...+99^2\)

\(A=1+2^2+3^2+4^2+5^2+...+99^2\)

\(A=1+2.\left(3-1\right)+3.\left(4-1\right)+...+99.\left(100-1\right)\)

\(A=\left(2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)\)

\(A=\frac{99.100.101}{3}-\frac{99.\left(99+1\right)}{2}\)

\(A=333300-4950=328350\)

11 tháng 6 2018

E=12+22+32+42+...+982+992+1002

=1+2(1+1)+3(1+2)+4(1+3)+....+98(1+97)+99(1+98)+100(1+99)

=1+1.2+2+3+2.3+4+3.4+....+98+97.98+99+98.99+100+99.100

=(1+2+3+4+...+100)+(1.2+2.3+3.4+...+99.100)

Đặt A=1+2+3+...+100=\(\frac{\left(100+1\right).100}{2}=5050\)

Đặt B=1.2+2.3+3.4+...+99.100

3B=1.2.3+2.3.3+....+99.100.3

3B=1.2.3+2.3.(4-1)+...+99.100.(101-98)

3B=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100

3B=99.100.101

=>B=\(\frac{99.100.101}{3}=333300\)

Vậy E=A+B=5050+333300=338350

6 tháng 9 2019

\(A=1+3+3^2+...+3^{100}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{101}\)

\(\Rightarrow3A-A=3^{101}-1\)

\(\Rightarrow A=\frac{3^{101}-1}{2}\)

17 tháng 2 2018

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 \(34x+17y⋮17\)

\(\Rightarrow2x+3y⋮17\)

14 tháng 8 2017

câu g) 

\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)

\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)

\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)

\(=\frac{12}{3}=4\)

14 tháng 8 2017

câu mình trả lời sai rồi thông cảm

25 tháng 7 2017

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 :

\(D=2^{100}-2^{99}+2^{98}-.....-2\) \(\Rightarrow2D=2^{101}-2^{100}+2^{99}-.....-2^2\)

\(\Rightarrow2D+D=2^{101}-2=3D\) \(\Rightarrow D=\frac{2^{101}-2}{3}\)

25 tháng 7 2017

\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(2A=1+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)-\frac{100}{3^{100}}\)

Ta thấy biểu thức trong dấu ngoặc nhỏ hơn 1/2 ( tự chứng minh ) nên 2A < 1 + 1/2 

\(\Rightarrow A< \frac{3}{4}\)

25 tháng 7 2017

\(C=1+3+3^2+3^3+...+3^{100}\)

\(3C=3+3^2+3^3+3^4+...+3^{101}\)

\(3C-C=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)

\(2C=3^{101}-1\)

\(C=\frac{3^{101}-1}{2}\)

27 tháng 9 2017

A=2^ 100 -2^ 99+2 ^98 -2 ^97+.....+2 ^2 -2

=>2A=2^ 101 -2 ^100+2^ 99 -2 ^98+.....+2^ 3 -2^ 2

=>2A+A=2 ^101 -2 ^100+2^ 99 -2^ 98+.....+2^ 3 -2 ^2+2^ 100 -2^ 99+2 ^98 -2^ 97+....+2 ^2 -2

=>3A=2^ 201 -2

=>A=\(\frac{2^{201}-2}{3}\)

B=3^ 100 -3^ 99+3^ 98 -3^ 97+....+3 ^2 -3+1

=>3B=3^ 101 -3 ^100+3 ^99 -3^ 98+...+3 ^3 -3^ 2+3

=>3B+B=3^ 101 -3^100+3^ 99 -3 ^98+...+3 ^3 -3 ^2+3+3 ^100 -3^ 99+3^ 98 -3^ 97+....+3 ^2 -3+1

=>4B=3 ^101+1

=>B=\(\frac{3^{101}+1}{4}\)

a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

\(\Rightarrow A+2A=2^{101}-2\)

  \(A\left(1+2\right)=2^{101}-2\)

  \(A.3=2^{101}-2\)

  \(A=\frac{2^{101}-2}{3}\)

b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)

\(3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2\)

\(\Rightarrow B+3B=3^{101}-3\)

\(B\left(1+3\right)=3^{101}-3\)

\(4B=3^{101}-3\)

   \(B=\frac{3^{101}-3}{4}\)