Answer: (i) \(S_N=\frac{1}{3}N(25N^2+90N+83)\).

(ii) \(T_N=\frac{1}{30}\left(\frac{1}{6}-\frac{1}{5N+6}\right)=\frac{N}{30(5N+6)}\).

(iii) \(\displaystyle \lim_{N\to\infty} \left(N^{-3}S_NT_N\right)=\frac{5}{18}\).

We work with the two sums separately.

(i) Expand the general term of \(S_N\):

\((5r+1)(5r+6)=25r^2+35r+6.\)

So

\(S_N=\sum_{r=1}^N (25r^2+35r+6)=25\sum_{r=1}^N r^2+35\sum_{r=1}^N r+6\sum_{r=1}^N 1.\)

Using the standard results

\(\sum_{r=1}^N r=\frac{N(N+1)}{2},\qquad \sum_{r=1}^N r^2=\frac{N(N+1)(2N+1)}{6},\qquad \sum_{r=1}^N 1=N,\)

we get

\(S_N=25\cdot \frac{N(N+1)(2N+1)}{6}+35\cdot \frac{N(N+1)}{2}+6N.\)

Put everything over 6:

\(S_N=\frac{25N(N+1)(2N+1)+105N(N+1)+36N}{6}.\)

Now expand:

\(25N(N+1)(2N+1)=25N(2N^2+3N+1)=50N^3+75N^2+25N,\)

so

\(S_N=\frac{50N^3+75N^2+25N+105N^2+105N+36N}{6}\)

\(=\frac{50N^3+180N^2+166N}{6}=\frac13 N(25N^2+90N+83).\)

This is the required result.

(ii) First rewrite the summand by partial fractions:

\(\frac{1}{(5r+1)(5r+6)}=\frac{A}{5r+1}+\frac{B}{5r+6}.\)

Equivalently, since the denominator differs by 5, a convenient choice is

\(\frac{1}{(5r+1)(5r+6)}=\frac{1}{5}\left(\frac{1}{5r+1}-\frac{1}{5r+6}\right),\)

because

\(\frac{1}{5}\left(\frac{1}{5r+1}-\frac{1}{5r+6}\right)=\frac{1}{5}\cdot \frac{5}{(5r+1)(5r+6)}.\)

Hence

\(T_N=\frac15\sum_{r=1}^N\left(\frac{1}{5r+1}-\frac{1}{5r+6}\right).\)

Now write out the first few terms:

\(T_N=\frac15\left(\frac16-\frac1{11}+\frac1{11}-\frac1{16}+\cdots+\frac{1}{5N+1}-\frac{1}{5N+6}\right).\)

Everything cancels except the first positive term and the last negative term, so

\(T_N=\frac15\left(\frac16-\frac{1}{5N+6}\right).\)

This can also be simplified to

\(T_N=\frac{1}{30}-\frac{1}{5(5N+6)}=\frac{N}{30(5N+6)}.\)

(iii) From part (i),

\(S_N=\frac13N(25N^2+90N+83).\)

From part (ii),

\(T_N=\frac{N}{30(5N+6)}.\)

Therefore

\(N^{-3}S_NT_N=N^{-3}\cdot \frac13N(25N^2+90N+83)\cdot \frac{N}{30(5N+6)}.\)

Simplify the powers of \(N\):

\(N^{-3}S_NT_N=\frac{25N^2+90N+83}{90(5N+6)N}.\)

Divide numerator and denominator by \(N\):

\(N^{-3}S_NT_N=\frac{25N+90+83/N}{90(5+6/N)}.\)

As \(N\to\infty\), this tends to

\(\frac{25}{90\cdot 5}=\frac{5}{18}.\)

So the limit is \(\frac{5}{18}\).