Cho S=1/5^2-2/5^3+…+99/5^100-100/5^101 CMR: S<1/36 29/10/2021 Bởi Madelyn Cho S=1/5^2-2/5^3+…+99/5^100-100/5^101 CMR: S<1/36
Giải thích các bước giải: Ta có: $S=\dfrac{1}{5^2}-\dfrac{2}{5^3}+…+\dfrac{99}{5^{100}}-\dfrac{100}{5^{101}}$ $\to 5S=\dfrac{1}{5}-\dfrac{2}{5^2}+…+\dfrac{99}{5^{99}}-\dfrac{100}{5^{100}}$ $\to 5S+S=\dfrac15-\dfrac1{5^2}-…-\dfrac1{5^{100}}-\dfrac{100}{5^{101}}$ $\to 6S=\dfrac15-(\dfrac1{5^2}+…+\dfrac1{5^{100}})-\dfrac{100}{5^{101}}$ Lại có: $P=\dfrac1{5^2}+…+\dfrac1{5^{100}}$ $\to 5P=\dfrac1{5}+…+\dfrac1{5^{99}}$ $\to 5P-P=\dfrac15-\dfrac1{5^{100}}$ $\to 4P=\dfrac15-\dfrac1{5^{100}}$ $\to P=\dfrac14(\dfrac15-\dfrac1{5^{100}})$ $\to 6S=\dfrac15-\dfrac14(\dfrac15-\dfrac1{5^{100}})-\dfrac{100}{5^{101}}$ $\to 6S=\dfrac15-\dfrac1{20}+\dfrac1{4\cdot 5^{100}}-\dfrac{20}{5^{100}}$ $\to 6S=\dfrac3{20}-\dfrac{79}{4\cdot 5^{100}}<\dfrac3{20}$ $\to S<\dfrac{1}{40}<\dfrac1{36}$ Bình luận
Giải thích các bước giải:
Ta có:
$S=\dfrac{1}{5^2}-\dfrac{2}{5^3}+…+\dfrac{99}{5^{100}}-\dfrac{100}{5^{101}}$
$\to 5S=\dfrac{1}{5}-\dfrac{2}{5^2}+…+\dfrac{99}{5^{99}}-\dfrac{100}{5^{100}}$
$\to 5S+S=\dfrac15-\dfrac1{5^2}-…-\dfrac1{5^{100}}-\dfrac{100}{5^{101}}$
$\to 6S=\dfrac15-(\dfrac1{5^2}+…+\dfrac1{5^{100}})-\dfrac{100}{5^{101}}$
Lại có:
$P=\dfrac1{5^2}+…+\dfrac1{5^{100}}$
$\to 5P=\dfrac1{5}+…+\dfrac1{5^{99}}$
$\to 5P-P=\dfrac15-\dfrac1{5^{100}}$
$\to 4P=\dfrac15-\dfrac1{5^{100}}$
$\to P=\dfrac14(\dfrac15-\dfrac1{5^{100}})$
$\to 6S=\dfrac15-\dfrac14(\dfrac15-\dfrac1{5^{100}})-\dfrac{100}{5^{101}}$
$\to 6S=\dfrac15-\dfrac1{20}+\dfrac1{4\cdot 5^{100}}-\dfrac{20}{5^{100}}$
$\to 6S=\dfrac3{20}-\dfrac{79}{4\cdot 5^{100}}<\dfrac3{20}$
$\to S<\dfrac{1}{40}<\dfrac1{36}$