Question

CMR: \(B=\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{n^3}< \dfrac{1}{12}\)( n ∈ N; n ≥ 3)

Answer 1

ta có \(\dfrac{1}{3^3}< \dfrac{1}{3^3-3}\)

\(\dfrac{1}{4^3}< \dfrac{1}{4^3-4}\)

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

\(\dfrac{1}{n^3}< \dfrac{1}{n^3-n}\)

=> \(\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+....+\dfrac{1}{n^3}< \dfrac{1}{3^3-3}+\dfrac{1}{4^3-4}+....+\dfrac{1}{n^3-n}\)=>\(B< \dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)đặt \(C=\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

C=\(\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+.....+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)C=\(\dfrac{1}{6}-\dfrac{1}{n\left(n+1\right)}\)

=> C<\(\dfrac{1}{6}\)

mà\(\dfrac{1}{6}< \dfrac{1}{4}\)

=> C<\(\dfrac{1}{4}\)

ta lại có B<C

=> B<\(\dfrac{1}{4}\) (đpcm)

Answer 2

mk bị nhầm rồi xin lỗi nha

Answer 3

chữa lại

C=\(\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

C=\(\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}+.....+\dfrac{1}{\left(n-1\right)}-\dfrac{1}{n\left(n+1\right)}\right)\)

C=\(\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{n\left(n+1\right)}\right)\)

C=\(\dfrac{1}{12}-\dfrac{1}{2n\left(n+1\right)}\)

=> C<\(\dfrac{1}{12}\)

mà B<C

=> B<\(\dfrac{1}{12}\) (đpcm)