Question:

Step 1: Express the given series in summation notation. The series is given by $$S = C_0 + \frac{C_1}{1 \cdot 3} + \frac{C_2}{2 \cdot 3} + \frac{C_3}{3 \cdot 3} + \cdots + \frac{C_n}{(n+1) \cdot 3}$$ This can be written as $$S = \sum_{r=0}^{n} \frac{C_r}{(r+1) \cdot 3}$$ Step 2: Utilize an integral identity for the term $\frac{1}{r+1}$. We know that $\frac{1}{r+1} = \int_0^1 x^r \, dx$. Substituting this into the sum, we get $$S = \sum_{r=0}^{n} \frac{C_r}{3} \int_0^1 x^r \, dx$$ Step 3: Interchange the summation and integral. Since the sum is finite, we can interchange the order of summation and integration: $$S = \frac{1}{3} \int_0^1 \left( \sum_{r=0}^{n} C_r x^r \right) \, dx$$ Step 4: Apply the Binomial Theorem. By the Binomial Theorem, the sum $\sum_{r=0}^{n} C_r x^r$ is equal to $(1+x)^n$. Substituting this into the expression for $S$: $$S = \frac{1}{3} \int_0^1 (1+x)^n \, dx$$ Step 5: Evaluate the definite integral. The integral $\int_0^1 (1+x)^n \, dx$ is evaluated as follows: $$ \int_0^1 (1+x)^n \, dx = \left[ \frac{(1+x)^{n+1}}{n+1} \right]_0^1 $$ $$ = \frac{(1+1)^{n+1}}{n+1} - \frac{(1+0)^{n+1}}{n+1} $$ $$ = \frac{2^{n+1}}{n+1} - \frac{1^{n+1}}{n+1} $$ $$ = \frac{2^{n+1}-1}{n+1} $$ Step 6: Substitute the integral result back into the expression for $S$. $$S = \frac{1}{3} \left( \frac{2^{n+1}-1}{n+1} \right)$$ $$S = \frac{2^{n+1}-1}{3(n+1)}$$