Câu 9.5: Tính nhanh \(B = {1 \over {15}} + {1 \over {35}} + {1 \over {63}} + {1 \over {99}} + {1 \over {143}}\)
\(B = {1 \over {15}} + {1 \over {35}} + {1 \over {63}} + {1 \over {99}} + {1 \over {143}}\)
\(\eqalign{
& = {1 \over 2}\left( {{2 \over {3.5}} + {2 \over {5.7}} + {2 \over {7.9}} + {2 \over {9.11}} + {2 \over {11.13}}} \right) \cr
& = {1 \over 2}\left( {{{5 – 3} \over {3.5}} + {{7 – 5} \over {5.7}} + {{9 – 7} \over {7.9}} + {{11 – 9} \over {9.11}} + {{13 – 11} \over {11.13}}} \right) \cr
& = {1 \over 2}\left( {{1 \over 3} – {1 \over 5} + {1 \over 5} – {1 \over 7} + {1 \over 7} – {1 \over 9} + {1 \over 9} – {1 \over {11}} + {1 \over {11}} – {1 \over {13}}} \right) \cr
& = {1 \over 2}.\left( {{1 \over 3} – {1 \over {13}}} \right) \cr
& = {1 \over 2}.{{10} \over {39}} = {5 \over {39}} \cr} \)
Câu 9.6: Tính nhanh \(C = {1 \over 2} + {1 \over {14}} + {1 \over {35}} + {1 \over {65}} + {1 \over {104}} + {1 \over {152}}\)
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\(C = {1 \over 2} + {1 \over {14}} + {1 \over {35}} + {1 \over {65}} + {1 \over {104}} + {1 \over {152}}\)
\(C = {2 \over 4} + {2 \over {28}} + {2 \over {70}} + {2 \over {130}} + {2 \over {208}} + {2 \over {304}}\)
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\(\eqalign{
& = {2 \over {1.4}} + {2 \over {4.7}} + {2 \over {7.10}} + {2 \over {10.13}} + {2 \over {13.16}} + {2 \over {16.19}} \cr
& = {2 \over 3}.\left( {{2 \over {1.4}} + {2 \over {4.7}} + {2 \over {7.10}} + {2 \over {10.13}} + {2 \over {13.16}} + {2 \over {16.19}}} \right) \cr
& = {2 \over 3}\left( {{{4 – 1} \over {1.4}} + {{7 – 4} \over {4.7}} + {{10 – 7} \over {7.10}} + {{13 – 10} \over {10.13}} + {{16 – 13} \over {13.16}} + {{19 – 16} \over {16.19}}} \right) \cr
& = {2 \over 3}.\left( {1 – {1 \over 4} + {1 \over 4} – {1 \over 7} + {1 \over 7} – {1 \over {10}} + {1 \over {10}} – {1 \over {13}} + {1 \over {13}} – {1 \over {16}} + {1 \over {16}} – {1 \over {19}}} \right) \cr
& = {2 \over 3}.\left( {1 – {1 \over {19}}} \right) \cr
& = {2 \over 3}.{{18} \over {19}} = {{12} \over {19}} \cr} \)
Câu 9.7: Chứng tỏ rằng \(D = {1 \over {{2^2}}} + {1 \over {{3^2}}} + {1 \over {{4^2}}} + … + {1 \over {{{10}^2}}} < 1\)
\(D = {1 \over {{2^2}}} + {1 \over {{3^2}}} + {1 \over {{4^2}}} + … + {1 \over {{{10}^2}}} < {1 \over {1.2}} + {1 \over {2.3}} + {1 \over {3.4}} + … + {1 \over {9.10}}\)
\(D< 1 – {1 \over 2} + {1 \over 2} – {1 \over 3} + … + {1 \over 9} – {1 \over {10}}\)
\(D< 1 – {1 \over {10}} = {9 \over {10}} < 1\)
