Question

$3C_0 + 3^2 \frac{C_1}{2} + 3^3 \frac{C_2}{3} + ... + 3^{n+1} \frac{C_n}{n+1}$ equals to

• A. $\frac{3^{n+1}-1}{n+1}$
• B. $\frac{3^{n+1}+1}{n+1}$
• C. $\frac{4^{n+1}-1}{n+1}$
• D. $\frac{4^{n+1}-1}{n+1}$

Answer 1

Answer: (C), (D)

$\frac{4^{n+1}-1}{n+1}$

Answer 2

The sum is $S = \sum_{r=0}^{n} 3^{r+1} \frac{\binom{n}{r}}{r+1}$. Consider the binomial expansion $(1+x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r$. Integrate from $0$ to $3$: $\int_0^3 (1+x)^n dx = \left[\frac{(1+x)^{n+1}}{n+1}\right]_0^3 = \frac{4^{n+1}-1}{n+1}$. Also, $\int_0^3 \sum_{r=0}^{n} \binom{n}{r} x^r dx = \sum_{r=0}^{n} \binom{n}{r} \int_0^3 x^r dx = \sum_{r=0}^{n} \binom{n}{r} \left[\frac{x^{r+1}}{r+1}\right]_0^3 = \sum_{r=0}^{n} \binom{n}{r} \frac{3^{r+1}}{r+1} = S$. Therefore, $S = \frac{4^{n+1}-1}{n+1}$.