Sequences & Series – Problem 1: Find the general formula of

Let $S_n$ be the sum of the first $n$ terms of the sequence $\{a_n\}$. Given $a_1=3$, and $$2S_n=a_{n+1}-3\quad(n\in\mathbb N^+).$$

(1) Find the general formula of $\{a_n\}$.

(2) If $b_n=na_n$, find the sum $T_n$ of the first $n$ terms of $\{b_n\}$.

(1) When $n=1$, $2a_1=a_2-3$, so $a_2=9$.

When $n\ge2$, from $$2S_n=a_{n+1}-3,\qquad 2S_{n-1}=a_n-3,$$ subtracting the two equations gives $$2(S_n-S_{n-1})=a_{n+1}-a_n\Rightarrow 2a_n=a_{n+1}-a_n,$$ so $a_{n+1}=3a_n$. Since $a_2=3a_1$, $\{a_n\}$ is a geometric sequence with first term $3$ and common ratio $3$, hence $$a_n=3^n.$$

(2) From $b_n=na_n=n\cdot3^n$, we get $$T_n=\sum_{k=1}^n k3^k.$$ Use the shifted-subtraction method: $$\begin{aligned} T_n&=1\cdot3+2\cdot3^2+\cdots+n\cdot3^n,\\ 3T_n&=1\cdot3^2+2\cdot3^3+\cdots+n\cdot3^{n+1}. \end{aligned}$$ Subtracting gives $$-2T_n=(3+3^2+\cdots+3^n)-n3^{n+1} =\frac{3(1-3^n)}{1-3}-n3^{n+1}.$$ Simplifying yields $$T_n=\frac{(2n-1)3^{n+1}+3}{4}.$$

(1) The sequence $\{a_n\}$ has general term $a_n=3^n$, so it is a geometric sequence with first term $3$ and common ratio $3$.

(2) For $b_n=na_n=n\cdot3^n$, the sum of the first $n$ terms is $$T_n=\frac{(2n-1)3^{n+1}+3}{4}.$$ This is obtained by applying the shifted-subtraction method to $\sum_{k=1}^n k3^k$ and then simplifying the result.