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b, B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2 \(\times\) B = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2 \(\times\) B + B = 1 - \(\dfrac{1}{2^{100}}\)
3B = ( 1 - \(\dfrac{1}{2^{100}}\))
B = ( 1 - \(\dfrac{1}{2^{100}}\)) : 3
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
A\(\times\) 3 = 3 + 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+ \(\dfrac{1}{3^{n-1}}\)
A \(\times\) 3 - A = 3 - \(\dfrac{1}{3^n}\)
2A = 3 - \(\dfrac{1}{3^n}\)
A = ( 3 - \(\dfrac{1}{3^n}\)) : 2
a) \(A=1+3+3^2+...+3^{100}\)
\(3A=3+3^2+3^3+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...-2^3+2^2-2+1\)
\(2B=2^{101}-2^{100}+2^{99}-2^{98}+...-2^4+2^3-2^2+2\)
\(B+2B=\left(2^{100}-2^{99}+...-2+1\right)+\left(2^{101}-2^{100}+...-2^2+2\right)\)
\(3B=2^{101}+1\)
\(B=\frac{2^{101}+1}{3}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(A=2A-A=2-\frac{1}{2^{2011}}=\frac{2^{2012}-1}{2^{2011}}\)
Nhầm
\(A=2A-A=2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)
A = 1 + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +.......+\(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
3\(\times\) A = 3 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+........+ \(\dfrac{1}{3^{n-1}}\)
3A - A = 3 + \(\dfrac{1}{3}\) - 1 - \(\dfrac{1}{3^n}\)
2A = \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)
A = ( \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)): 2
A = \(\dfrac{7.3^{n-1}-1}{3^n}\) : 2
A = \(\dfrac{7.3^{n-1}-1}{2.3^n}\)
B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+......+\(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2B = 2 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2B + B = 2 - \(\dfrac{1}{2^{100}}\)
3B = 2 - \(\dfrac{1}{2^{100}}\)
B = ( 2 - \(\dfrac{1}{2^{100}}\)): 3
B = \(\dfrac{2.2^{100}-1}{2^{100}}\) : 3
B = \(\dfrac{2^{101}-1}{3.2^{100}}\)
A = 1 + 1/22+1/23+...+1/22015
(1-1/2) A = (1-1/2) (1+1/22+1/23+...+1/22015) = 1 - 1/22016
A = 2 *( 1 -1/22016) = 2 -1/22015
A = 1 + 1/22+1/23+...+1/22015
(1-1/2) A = (1-1/2) (1+1/22+1/23+...+1/22015) = 1 - 1/22016
A = 2 *( 1 -1/22016) = 2 -1/22015
\(\frac{\left(a^3+a^2\right)+\left(a^2-1\right)}{\left(a^3+a^2\right)+\left(a^2+a\right)+\left(a+1\right)}=\frac{a^2\left(a+1\right)+\left(a+1\right)\left(a+1\right)}{a^2\left(a+1\right)+a\left(a+1\right)+\left(a+1\right)}=\frac{\left(a+1\right)\left(a^2+a-1\right)}{\left(a+1\right)\left(a^2+a+1\right)}=\frac{a^2+a-1}{a^2+a-1}\)
a) \(A=\left(-m+n-p\right)-\left(-m-n-p\right)\)
\(=-m+n-p+m+n+p=2n\)
Vậy A=2n
b) Ta có A=2n
Thay số vào ta được A=2.(-1)=-2
Vậy A=-2 khi n=-1
1+1/A+1/a2+1/a3+1+.../an+1
=1(1/A/a2/a3/...an)
=1.(1/a1+2+3+...+n)
=1.(1/a6+...+n)
=a6+...+n