Sum of series of product of terms

The $n$-th term of the series is $T_n = (4n + 3)(4n + 7)$ for $n = 1, 2, \dots, 23$. (Since $4(23) + 3 = 95$).

We use the telescoping identity: $T_n = \frac{1}{12} [(4n+3)(4n+7)(4n+11) - (4n-1)(4n+3)(4n+7)]$ Sum $S = \sum_{n=1}^{23} T_n = \frac{1}{12} [(95)(99)(103) - (3)(7)(11)]$ $S = \frac{1}{12} [968685 - 231] = \frac{968454}{12} = 80704.5$

Wait! Let's check $T_n$ formula accurately: $T_1 = 7 \times 11$, $T_{23} = 95 \times 99$. Difference between terms is 4. Formula for sum of $a_n a_{n+1}$ where $a_n = 4n+3$: $S = \sum_{n=1}^{23} (16n^2 + 40n + 21)$ $\sum n^2 = \frac{23 \times 24 \times 47}{6} = 4324$ $\sum n = \frac{23 \times 24}{2} = 276$ $S = 16(4324) + 40(276) + 21(23) = 69184 + 11040 + 483 = 80707$

Thus, $S = 80707$.