Question

chứng minh rằng:1/2²+1/3²+1/4²+......+1/100² < 1

 

Answer 1

Ta có :

1/2² < 1/1.2 = 1/1 - 1/2

1/3² < 1/2.3 = 1/2 - 1/3

..........

1/100² < 1/99.100 = 1/99 - 1/100

=> 1/2²+1/3²+1/4²+......+1/100² <  1/1 - 1/2 + 1/2 - 1/3 + .... + 1/99 - 1/100 = 1 - 1/100 < 1

=>  1/2²+1/3²+1/4²+......+1/100² < 1 ( dpcm )

Answer 2

Gọi tổng trên bằng A. Ta có

A<1/10^2 x 100=1

Nên A<1

Answer 3

toan lop 5 ma co so mu roi ak

Answer 4

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}\)

.......

\(\frac{1}{100^2}< \frac{1}{99.100}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< 1-\frac{1}{100}< 1\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< 1\)

=> Đpcm

Answer 5

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