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\(\frac{a_1-1}{100}=\frac{a_2-2}{99}=\frac{a_3-3}{98}=...=\frac{a_{100}-100}{1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a_1-1+a_2-2+a_3-3+...+a_{100}-100}{1+2+3+...+100}\)\(=\)\(\frac{a_1+a_2+a_3+...+a_{100}-\left(1+2+3+...+100\right)}{1+2+3+...+100}\)
\(=\)\(\frac{10100-5050}{5050}\)vì \(1+2+3+...+100=5050\)
\(=\) \(\frac{5050}{5050}\)\(=\)\(1\)
Ta có \(\frac{a_1-1}{100}=1\Rightarrow a_1-1=100\Rightarrow a_1=101\)
\(\frac{a_2-2}{99}=1\Rightarrow a_2-2=99\Rightarrow a_2=101\)
\(\frac{a_3-3}{98}=1\Rightarrow a_3-3=98\Rightarrow a_3=101\)
\(....\)
\(\frac{a_{100}-100}{1}=1\Rightarrow a_{100}-100=1\Rightarrow a_{100}=101\)
Vậy \(a_1=a_2=a_3=....=a_{100}=101\)
Giải:
a) \(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)
\(\Leftrightarrow3A=3+3^2+3^3+3^4+...+3^{100}+3^{101}\)
\(\Leftrightarrow3A-A=2A=3^{101}-1\)
\(\Leftrightarrow A=\dfrac{3^{101}-1}{2}\)
b) \(B=1-3+3^2+3^3+...+3^{99}+3^{100}\)
\(\Leftrightarrow3B=3-3^2+3^3+3^4+...+3^{100}+3^{101}\)
\(\Leftrightarrow3B-B=2B=3^{101}-1-6-18=3^{101}--25\)
\(\Leftrightarrow B=\dfrac{3^{101}-25}{2}\)
Chúc bạn học tốt!
\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)
\(3A=3+3^2+3^3+3^4+...+3^{100}+3^{101}\)
\(3A-A=\left(3+3^2+3^3+3^4+...+3^{100}+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{99}\right)\)
\(2A=3^{101}-1\Leftrightarrow A=\dfrac{3^{101}-1}{2}\)
B đề sai
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)
\(A=1-\frac{1}{2^{99}}\)
C= 1/100-(1/1.2+1/2.3+...+1/97.98+1/98.99+1/99.100)
C=1/100-(1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)
C=1/100-(1-1/100)
C=1/100-99/100
C=-98/100=-49/50
\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)+\dfrac{1}{100}\)
\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=\left(-1\right)+\dfrac{1}{50}=-\dfrac{49}{50}\)
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 \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
Đặt A = 2 100 + 2 99 + 2 98 + .... + 22 +2 + 1
2A = 2 101 + 2 100 + 2 99 + .... + 2 3 + 2 2 + 2
2A - A = ( 2 101 + 2 100 + 2 99 + .... + 2 3 + 2 2 + 2 )
- ( 2 100 + 2 99 + 2 98 + .... + 22 +2 + 1 )
A = 2 101 - 1
= 2101 + 2100 + 299 + ... + 22 + 2 + 1
=(2101 - 1): 2