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Xét \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{123}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{122}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-2\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{61}\right)\)
\(=\frac{1}{62}+\frac{1}{63}+\frac{1}{64}+...+\frac{1}{123}\)
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A = \(\dfrac{3^{123}+1}{3^{125}+1}\) Vì 3123 + 1 < 2125 + 1 Nên A = \(\dfrac{3^{123}+1}{3^{125}+1}\)< \(\dfrac{3^{123}+1+2}{3^{125}+1+2}\)
A < \(\dfrac{3^{123}+3}{3^{125}+3}\) = \(\dfrac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\) = \(\dfrac{3^{122}+1}{3^{124}+1}\) = B
Vậy A < B
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Đáp án cần chọn là: D
M = 1 + 1 2 + 1 2 2 + 1 2 3 + ... + 1 2 99 + 2 2 100 2 M = 2. 1 + 1 2 + 1 2 2 + 1 2 3 + ... + 1 2 99 + 2 2 100 = 2 + 1 + 1 2 + 1 2 2 + 1 2 3 + ... + 1 2 99
Ta có :
M = 2 M − M = ( 2 + 1 + 1 2 + 1 2 2 + 1 2 3 + ... + 1 2 99 ) − ( 1 + 1 2 + 1 2 2 + 1 2 3 + ... + 1 2 99 + 2 2 100 ) M = 2 M − M = 2 + 1 + 1 2 + 1 2 2 + 1 2 3 + ... + 1 2 98 + 1 2 99 − 1 + 1 2 + 1 2 2 + 1 2 3 + ... + 1 2 99 + 1 2 100 = 2 − 1 2 100 = 2 101 − 1 2 100
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a)\(...A=\dfrac{2^{50+1}-1}{2-1}=2^{51}-1\)
b) \(...\Rightarrow B=\dfrac{3^{80+1}-1}{3-1}=\dfrac{3^{81}-1}{2}\)
c) \(...\Rightarrow C+1=1+4+4^2+4^3+...+4^{49}\)
\(\Rightarrow C+1=\dfrac{4^{49+1}-1}{4-1}=\dfrac{4^{50}-1}{3}\)
\(\Rightarrow C=\dfrac{4^{50}-1}{3}-1=\dfrac{4^{50}-4}{3}=\dfrac{4\left(4^{49}-1\right)}{3}\)
Tương tự câu d,e,f bạn tự làm nhé
1 - 1/2 + 1/3 - 1/4 + .. + 1/121 - 1/122 + 1/123
= ( 1 + 1/3 + ... + 1/121 + 1/123 ) - ( 1/2 + 1/4 + .. + 1/122 )
Đến bước này ta sẽ cùng cộng 2 vế với : 1/2 + 1/4 + .. + 1/122
= ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/121 + 1/122 + 1/123 ) - 2. ( 1/2 + 1/4 + .. + 1/122 )
= ( 1 + 1/2 + 1/3 + ...+ 1/122 +1 /123 ) - ( 1 + 1/2 + ...+ 1/61 )
= 1/62 + 1/63 + ..+1/122 + 1/123
Chúc học giỏi !!