Precalculus, Chapter 9, 9.4, Section 9.4, Problem 43

1/4,1/12,1/24,1/40,.......1/(2n(n+1))
Let's write down the first few sums of the sequence,
S_1=1/4
S_2=1/4+1/12=(3+1)/12=4/12=1/3=2/(2(2+1))
S_3=1/4+1/12+1/24=(6+2+1)/24=9/24=3/8=3/(2(3+1))
S_4=1/4+1/12+1/24+1/40=(30+10+5+3)/120=48/120=2/5=4/(2(4+1))
From the above , it appears that the formula for the sum of the k terms of the sequence is,
S_k=k/(2(k+1))
We have verify the above the formula for k=1 by plugging in k=1
S_1=1/(2(1+1))=1/4
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)=1/4+1/12+1/24+1/40+......+1/(2k(k+1))+1/(2(k+1)(k+1+1))
S_(k+1)=k/(2(k+1))+1/(2(k+1)(k+2))
S_(k+1)=(k(k+2)+1)/(2(k+1)(k+2))
S_(k+1)=(k^2+2k+1)/(2(k+1)(k+2))
S_(k+1)=((k+1)(k+1))/(2(k+1)(k+2))
S_(k+1)=(k+1)/(2(k+2))
S_(k+1)=(k+1)/(2(k+1+1))
So the formula is true for n=k+1 also,
Hence the formula for the sum of n terms of the sequence is,
S_n=n/(2(n+1))