cho A= 7/4 + 17/9 + 31/16 + …… 4999/2500 chứng minh A <97 27/10/2021 Bởi Aubrey cho A= 7/4 + 17/9 + 31/16 + …… 4999/2500 chứng minh A <97
TL: $A =\frac{7}{4}$ $+ \frac{17}{9}$ $+ …+\frac{4999}{2500}$ $A =( 1+ \frac{3}{4})( 1+ \frac{8}{9})+…+( 1+ \frac{2499}{2500}) ( 49 $ $nhóm )$ $A = 1+ \frac{3}{4} 1+ \frac{8}{9}+…+ 1+ \frac{2499}{2500}$ $A = ( 1 +…+1) + $ $( \frac{3}{4}$ $+\frac{8}{9}+ …+$ $\frac{2499}{2500})$ $A = 49 + $ $ ( \frac{3}{4}$ $+\frac{8}{9}+ …+$ $\frac{2499}{2500})$ Đặt B = $ \frac{3}{4}$ $+\frac{8}{9}+ …+$ $\frac{2499}{2500}$ B = $1-\frac{1}{2^2} + 1 – $ $\frac{1}{3^2} + … + $ $1 – \frac{1}{50^2}$ $B = 1 + … + 1 -$ $( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ … + $ $\frac{1}{50^2})$ $B = 49 – ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ … + $ $\frac{1}{50^2})$ Thay B ta có: $A = 49 + B$ $A = 49 + 49 – ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ … + $ $\frac{1}{50^2})$ $ A = 98 – ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ … + $ $\frac{1}{50^2})$ $A = 97 + [ 1 – ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ … + $ $\frac{1}{50^2}) > 97 $ $=> A > 97$ Bình luận
TL:
$A =\frac{7}{4}$ $+ \frac{17}{9}$ $+ …+\frac{4999}{2500}$
$A =( 1+ \frac{3}{4})( 1+ \frac{8}{9})+…+( 1+ \frac{2499}{2500}) ( 49 $ $nhóm )$
$A = 1+ \frac{3}{4} 1+ \frac{8}{9}+…+ 1+ \frac{2499}{2500}$
$A = ( 1 +…+1) + $ $( \frac{3}{4}$ $+\frac{8}{9}+ …+$ $\frac{2499}{2500})$
$A = 49 + $ $ ( \frac{3}{4}$ $+\frac{8}{9}+ …+$ $\frac{2499}{2500})$
Đặt B = $ \frac{3}{4}$ $+\frac{8}{9}+ …+$ $\frac{2499}{2500}$
B = $1-\frac{1}{2^2} + 1 – $ $\frac{1}{3^2} + … + $ $1 – \frac{1}{50^2}$
$B = 1 + … + 1 -$ $( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ … + $ $\frac{1}{50^2})$
$B = 49 – ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ … + $ $\frac{1}{50^2})$
Thay B ta có:
$A = 49 + B$
$A = 49 + 49 – ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ … + $ $\frac{1}{50^2})$
$ A = 98 – ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ … + $ $\frac{1}{50^2})$
$A = 97 + [ 1 – ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ … + $ $\frac{1}{50^2}) > 97 $
$=> A > 97$