It is easy enough, even with that incorrect formula :)
The correct formula should be n∑i=1i2=n(n+1)(2n+1)6.
Obviously, this limit is a Riemann integral.
S=limn→∞n∑i=1((2n)(in)2)=2limn→∞n∑k=1((1n)(kn)2)Riemann integral: For any function f continuous on [0,1], limn→∞n∑k=11nf(kn)=∫10f(x)dxS=2∫10x2dx=2(13)=23