Precalculus, Chapter 9, 9.4, Section 9.4, Problem 44

1/(2*3),1/(3*4),1/(4*5),1/(5*6),..........1/((n+1)(n+2))
Let's write down the first few sums of the sequence,
S_1=1/(2*3)=1/6=1/(2(1+2))
S_2=1/(2*3)+1/(3*4)=1/6+1/12=(2+1)/12=3/12=1/4=2/(2(2+2))
S_3=1/(2*3)+1/(3*4)+1/(4*5)=1/4+1/20=(5+1)/20=3/10=3/(2(3+2))
S_4=1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)=3/10+1/30=(9+1)/30=1/3=4/(2(4+2))
From the above, it appears that the formula for the k terms of the sequence is,
S_k=k/(2(k+2))
The formula is already verified for n=1,
Now let's assume that the formula is valid for n=k and we have to show that it is valid for n=k+1
S_(k+1)=1/(2*3)+1/(3*4)+.........1/((k+1)(k+2))+1/((k+2)(k+3))
S_(k+1)=k/(2(k+2))+1/((k+2)(k+3))
S_(k+1)=(k(k+3)+2)/(2(k+2)(k+3))
S_(k+1)=(k^2+3k+2)/(2(k+2)(k+3))
S_(k+1)=(k^2+2k+k+2)/(2(k+2)(k+3))
S_(k+1)=(k(k+2)+1(k+2))/(2(k+2)(k+3))
S_(k+1)=((k+2)(k+1))/(2(k+2)(k+3))
S_(k+1)=(k+1)/(2(k+3))
S_(k+1)=(k+1)/(2(k+1+2))
So the formula is true for n=k+1 also,
Hence the formula for the sum of the n terms of the sequence is,
S_n=n/(2(n+2))