Question

1.Tính tổng :

S = \(\dfrac{3}{1.2}\) + \(\dfrac{3}{2.3}\) + \(\dfrac{3}{3.4}\) + \(\dfrac{3}{4.5}\) + ...... + \(\dfrac{3}{2015.2016}\)

Mong các bạn giúp mình nhé cái này quan trọng đối với mình lắm

Answer 1

S = \(\dfrac{3}{1.2}\)+\(\dfrac{3}{2.3}\)+\(\dfrac{3}{3.4}\)+\(\dfrac{3}{4.5}\)+...+\(\dfrac{3}{2015.2016}\)

= 3.\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2015.2016}\right)\)

= 3.\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)

= 3.\(\left(1-\dfrac{1}{2016}\right)\) = 3.\(\dfrac{2015}{2016}\)=\(\dfrac{3.2015}{2016}\)=\(\dfrac{1.2015}{672}\)=\(\dfrac{2015}{672}\)

Vậy S = \(\dfrac{2015}{672}\)

Answer 2

Ta có S=\(\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+\dfrac{3}{4.5}+...+\dfrac{3}{2015.2016}\)

=3.(\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2015.2016}\))

=3.(\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\))

=\(3.\left(1-\dfrac{1}{2016}\right)\)

= \(3-\dfrac{1}{672}\)=\(\dfrac{2015}{672}=2\dfrac{671}{672}\)

Answer 3

Ta có \(S=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\right)\)
\(S=3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(S=3\left(1-\dfrac{1}{2016}\right)\)
\(S=\dfrac{3.2015}{2016}\)
\(S=\dfrac{6045}{2016}\)