For the given series:-
\(T_n = \frac{n^2 + 6n +12}{n(n + 1)(n + 2)} \times \frac 1{2^{n + 1}}\)
\(S = \sum\limits_{i = 1}^n \frac{i^2 + 6i + 12}{i(i +1)(i + 2)} \times \frac1{2^{i + 1}}\)
\(S = \sum \limits_{i = 1}^n \frac{i + 3}{i(i + 1)} \times \frac1{2^i} - \frac{i + 4}{(i + 1)(i + 2)2^{i + 1}} \)
Middle term will cancel out
\(S = 1 - \frac{n + 4}{(n + 1)(n + 2)} \times \frac1{2^{n + 1}}\)
Hence, sum = \( 1 - \frac{n + 4}{(n + 1)(n + 2)} \times \frac1{2^{n + 1}}\)
