Question

Evaluate the following

1/(1*2)  + 1/(2*3)   +  1/(3*4)  +  1/(4*5) + ... + 1/(98*99) + 1/(99*100)

Answer 1

99/100

Take the first four terms: 1/2+1/6+1/12+1/20=4/5.

In general, the first n terms sum to n/(n+1).

To prove by induction, assume S[n]=n/(n+1) then

S[n+1]=n/(n+1)+1/((n+1)(n+2))=

(n(n+2)+1)/((n+1)(n+2))=

(n²+2n+1)/((n+1)(n+2))=

(n+1)²/((n+1)(n+2))=(n+1)/(n+2)=S[n+1]

When n=1 (base case), S₁=½ and S₂=1/(1×2)+1/(2×3)=½+⅙=⅔. And we expect this using the formula.

In the question n=99 so S₉₉=99/100.