The sum $\frac{19}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{28}{2 \cdot 3 \cdot 4 \cdot 8} + \frac{39}{3 \cdot 4 \cdot 5 \cdot 16} + \frac{52}{4 \cdot 5 \cdot 6 \cdot 32} + ... $ upto infinite terms is equal to:
Step-by-Step Solution
Key Concept: Expressing the numerator as a linear combination of products in the denominator enables telescoping of the partial fraction sum.
The sum $S_n = \sum_{r=1}^{n} \frac{r^2 + 6r + 12}{(r)(r+1)(r+2)2^{r+1}} = \sum_{r=1}^{n} \frac{2(r+2)(r+3) - r(r+4)}{(r)(r+1)(r+2)2^{r+1}}$ decomposes into partial fractions. After telescoping, we obtain $S_n = 1 - \frac{n+4}{(n+1)(n+2)2^{n+1}}$.
Correct Answer: 2