If ∑r=1n tr=112n(n+1)(n+2), the value ∑r=1n 1tr is
2nn+1
n−1(n+1)!
4nn+1
3nn+2
We have tr=∑k=1r tk−∑k=1r−1 tk
=112r(r+1)(r+2)−112(r−1)(r)(r+1)=14r(r+1)
Now, 1tr=4r(r+1)=41r−1r+1
⇒ ∑r=1n 1tr=4∑r=1n 1r−1r+1=41−1n+1=4nn+1