Sequences & Series – Problem 3: Find the general formula of

Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence $\{a_n\}$, and define $$b_n=\begin{cases} a_n-301,& n\text{ is odd},\\ 2^{a_n},& n\text{ is even}. \end{cases}$$ Given $b_4=32$ and $S_5=20$.

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

(2) Let $T_n$ be the sum of the first $n$ terms of $\{b_n\}$, and define $c_n=2T_{2n}-S_{2n}$.

(i) Find an expression for $c_n$.

(ii) If integer $n$ satisfies $c_n\le0$, find the maximum value of $n$, and justify your answer.

(1) Let the first term of the arithmetic sequence $\{a_n\}$ be $a_1$, and common difference be $d$. From $b_4=32$, we get $2^{a_4}=32\Rightarrow a_4=5$. From $S_5=20$, we get $\dfrac{5(a_1+a_5)}2=20\Rightarrow a_3=4$. Solving $$a_3=a_1+2d=4,\qquad a_4=a_1+3d=5$$ gives $a_1=2,d=1$, hence $$a_n=n+1.$$

(2) From $a_n=n+1$, we have $$b_n=\begin{cases} n-300,& n\text{ is odd},\\ 2^{n+1},& n\text{ is even}. \end{cases}$$ Also, $$S_{2n}=n(2n+3).$$ And $$T_{2n}=\sum_{k=1}^{n}(2k-301)+\sum_{k=1}^{n}2^{2k+1}=n(n-300)+\frac{2^{2n+3}-8}{3}.$$ Therefore $$c_n=2T_{2n}-S_{2n}=\frac{4^{n+2}-16}{3}-603n.$$

(ii) $$c_n-c_{n-1}=4^{n+1}-603.$$ When $n\ge4$, $c_n-c_{n-1}>0$, so $\{c_n\}$ is increasing afterward. Direct computation: $$c_4=\frac{4^6-16}{3}-603\times4=\frac{4096-16}{3}-2412=\frac{4080}{3}-2412=1360-2412=-1052<0,$$ $$c_5=\frac{4^7-16}{3}-603\times5=\frac{16384-16}{3}-3015=\frac{16368}{3}-3015=5456-3015=2441>0.$$ Hence the largest integer satisfying $c_n\le0$ is $$n=4.$$

(1) The arithmetic sequence is $a_n=n+1$, obtained from $b_4=32$ and $S_5=20$, which determine $a_1=2$ and $d=1$.

(2)(i) Using the parity definition of $b_n$, the expression for $c_n$ is $$c_n=\frac{4^{n+2}-16}{3}-603n.$$ (ii) Since $c_n-c_{n-1}=4^{n+1}-603$ becomes positive from $n=4$ onward and $c_4<0<c_5$, the maximum integer with $c_n\le0$ is $n=4$.