Cho A = $\frac{3^3}{1}$ – $\frac{5^3}{3}$ + $\frac{7^3}{6}$ – $\frac{9^3}{10}$ + $\frac{11^3}{15}$ – $\frac{13^3}{21}$ + $\frac{15^3}{25}$ – $\frac{17^3}{36}$ +…+ $\frac{199^3}{4950}$ . So sánh A với 814
Cho A = $\frac{3^3}{1}$ – $\frac{5^3}{3}$ + $\frac{7^3}{6}$ – $\frac{9^3}{10}$ + $\frac{11^3}{15}$ – $\frac{13^3}{21}$ + $\frac{15^3}{25}$ – $\frac{17^3}{36}$ +…+ $\frac{199^3}{4950}$ . So sánh A với 814
$\frac{A}{2}$ = $\frac{3^3}{2}$ – $\frac{5^3}{6}$ + $\frac{7^3}{12}$ – $\frac{9^3}{20}$ + $\frac{11^3}{30}$ – $\frac{13^3}{42}$ + $\frac{15^3}{56}$ – $\frac{17^3}{72}$ + … + $\frac{199^3}{9900}$
= 3² . ( 1 + $\frac{1}{2}$ ) – 5² . ( $\frac{1}{2}$ + $\frac{1}{3}$ ) + 7² . ( $\frac{1}{3}$ + $\frac{1}{4}$ ) – 9² . ( $\frac{1}{4}$ + $\frac{1}{5}$ ) + … + 199² . ( $\frac{1}{99}$ + $\frac{1}{100}$ )
= 3² + ( $\frac{3^2}{2}$ – $\frac{5^2}{2}$ ) – ( $\frac{5^2}{3}$ – $\frac{7^2}{3}$ ) + ( $\frac{7^2}{4}$ – $\frac{9^2}{4}$ ) – ( $\frac{9^2}{5}$ – $\frac{11^2}{5}$ ) + … + ( $\frac{197^2}{99}$ – $\frac{199^2}{99}$ ) + $\frac{199^2}{100}$
= 3² – 8+8-8+…+8 + $\frac{199^2}{100}$ = 3² + $\frac{199^2}{100}$ < 3² + $\frac{199.200}{100}$ = 9 + 398 = 407
⇒ A < 407 . 2 = 814
Ta có:
`A=3^3/1-5^3/3+7^3/6-9^3/10+11^3/15-13^3/21+15^3/25-17^336+…+199^3/4950`
`⇒A/2=3^3/2-5^3/6+7^3/12-9^3/20+11^3/30-13^3/42+…+199^3/9900`
`A/2=3^3/1.2-5^3/2.3+7^3/3.4-9^3/4.5+11^3/5.6-13^3/6.7+…+199^3/99.100`
`A/2=3^3(1-1/2)-5^3(1/2-1/3)+7^3(1/3-1/4)-9^3(1/4-1/5)+…+199^3(1/99-1/100)`
`A/2=3^3-(3^3+5^3)/2+(5^3+7^3)/3-(7^3+9^3)/4+…+(197^3+199^3)/99-199^3/100`
`A/2=3^3-199^3/100-(16.2^2+12)+(16.3^2+12)-(16.4^2+12)+…+(16.99^2+12)`
`A/2=3^3-199^3/100+16(3^2-2^2+5^2-4^2+7^2-6^2+…+99^2-98^2)`
`A/2=3^3-199^3/100+16(2+3+4+5+…+98+99)`
`A/2=3^3-199^3/100+16(99.50-1)`
`⇒A=16.99.100-199^3/50+22`
`⇒A=(2^3.100^2(100-1)-199^3)/50+22`
`=(200^3-199^3-2.200^2)/50+22`
`=(200^2 + 200.199 + 199^2 − 2.200^2)/50+22`
`=(199^2 − 200^2 + 200.199)/50+22`
`=(− 199 − 200 + 200.199)/50+22=199^2/50+18`
`⇒A=199^2/50+18<(199.200)/50+18=814`
Vậy `A<814`.