Let Tr be the rth term of the given series. Then,
Tr=(2r+1)r2(r+1)2,
=[1r2−1(r+1)2]
Hence, the required sum is
n∑r=1Tr=n∑r=1[1r2−1(r+1)2]
=(112−122)+(122−132)+(⋮ ⋮)+(1n2−1(n+1)2)
=1−1(n+1)2
=2n+n2(n+1)2
Now,
0.99=2n+n2(n+1)2
⇒99100=2n+n2(n+1)2⇒99n2+99+198n=200n+100n2⇒n2+2n−99=0⇒n=−11,9
But n∈N
∴n=9