Question

[Trần Thu Hà]

Tính:
S₁=1+3+5+...+(2n+1) (n: số nguyên cho trước).
S₂=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{n}\) (n: số nguyên cho trước).
S₃=\(\dfrac{1}{a}\)+\(\dfrac{1}{a+1}\)+\(\dfrac{1}{a+2}\)+...+\(\dfrac{1}{a+100}\).