Question

p= 1/2^2 +1/3^2 +1/4^2+....+ 1/100^2

dấu / là phân số,còn ^ là mũ

giúp em vs!!!!

Answer 1

đề bài là j

Answer 2

vì \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

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

.............

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

=>P=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...\dfrac{1}{99.100}\)

Mà \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)

=\(1-\dfrac{1}{100}\)<1

=>P<1 (đpcm)