Question

\(VT=\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{\left(4n-3\right)\left(4n+1\right)}\)

\(VT=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(VT=\dfrac{1}{4}\left(1-\dfrac{1}{4n+1}\right)\)

\(VT=\dfrac{1}{4}\left(\dfrac{4n}{4n+1}\right)\)

\(VT=\dfrac{4n}{16n+4}=\dfrac{4n}{4\left(4n+1\right)}=\dfrac{n}{4n+1}=VP\)

Xảy ra với mọi \(n\in Z^+\)

@Đặng Thị Cẩm Tú

Answer 1

Cái này là j

Answer 2

- Không hiểu gì hết