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A= \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)
2A= \(2.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\right)\)
2A= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
⇒ 2A- A= \(1-\dfrac{1}{2^{100}}\)
⇒ A= \(1-\dfrac{1}{2^{100}}\)
B= \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
3B= \(3.\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\)
3B= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
⇒ 3B- B= \(1-\dfrac{1}{3^{100}}\)
⇒ B.(3-1)= \(1-\dfrac{1}{3^{100}}\)
⇒ 2B= \(1-\dfrac{1}{3^{99}}\)
⇒ B= \(\left(1-\dfrac{1}{3^{99}}\right):2\)
⇒ B= \(\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)
a>
\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000
ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )
1/100^2<1/2
=>A<1
a)( 100 - 1^2 ) * ( 100 - 2^2 ) * ( 100 - 3^2 ) * ...... * ( 100 -50^2 )=( 100 - 1^2 ) * ( 100 - 2^2 ) * ( 100 - 3^2 ) * ...... *(100-10^2)....* ( 100 -50^2 )=( 100 - 1^2 ) * ( 100 - 2^2 ) * ( 100 - 3^2 ) * ...... *(0)....* ( 100 -50^2 )=0
b)1^0 + 1^2 + 1^3+ 1^4 +..........+1^99=1+1+1+1+....+1+1+1(có 100 số 1)=100x1=100
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
\(\Rightarrow\)\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)
\(\Rightarrow\)\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow\)\(A=2-\frac{1}{2^{100}}\)
\(B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow\)\(3B=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\)
\(\Rightarrow\)\(3B-B=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow\)\(2B=3-\frac{1}{3^{100}}\)
\(\Rightarrow\)\(B=\frac{3-\frac{1}{3^{100}}}{2}\)
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)
\(\frac{1}{2}B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{101}\)
\(\frac{1}{2}B-B=\left(\frac{1}{2}\right)^{101}-\frac{1}{2}\)
\(\frac{-1}{2}B=\left(\frac{1}{2}\right)^{101}-\frac{1}{2}\)
\(B=\frac{\left(\frac{1}{2}\right)^{101}-\frac{1}{2}}{-\frac{1}{2}}\)