Sequences & Series – Problem 4: find the value of

Inserting the sum of any two adjacent terms into a sequence is called one "sum-growth" operation; inserting the product of the two adjacent terms is called one "product-growth" operation. Starting from the sequence $1,4$: one sum-growth gives $1,5,4$, and two sum-growth operations give $1,6,5,9,4$; one product-growth gives $1,4,4$, and two product-growth operations give $1,4,4,16,4$. After $n$ sum-growth operations, the sequence is $1,x_1,x_2,\ldots,x_k,4$; after $n$ product-growth operations, the sequence is $1,y_1,y_2,\ldots,y_k,4$. Define $$a_n=1+x_1+x_2+\cdots+x_k+4,\quad b_n=\log_4(1\cdot y_1y_2\cdots y_k\cdot4).$$

(1) When $n=4$, find the value of $k$.

(2) Prove that the sequence $\left\{a_n-\dfrac52\right\}$ is geometric.

(3) Find the sum of the first $n$ terms of the sequence $\{n(a_n+b_n)\}$.

(1) Let $c_n$ be the number of inserted terms after the $n$-th product-growth operation. Then $k=c_n$. There are $c_n+1$ gaps, so after the $(n+1)$-th product-growth operation, $c_n+1$ new terms are inserted, hence $$c_{n+1}=c_n+(c_n+1)=2c_n+1.$$ Thus $c_{n+1}+1=2(c_n+1)$, and $c_1+1=2$, so $\{c_n+1\}$ is geometric with first term $2$ and ratio $2$: $$c_n+1=2^n\Rightarrow c_n=2^n-1.$$ Therefore $k=2^n-1$, and when $n=4$, $k=15$.

(2) Let $a_n$ be the sum of all terms after the $n$-th sum-growth operation. After the $(n+1)$-th sum-growth operation, the sum of newly inserted terms is $2a_n-(1+4)=2a_n-5$, so $$a_{n+1}=a_n+(2a_n-5)=3a_n-5,$$ i.e. $$a_{n+1}-\frac52=3\left(a_n-\frac52\right).$$ Hence $\left\{a_n-\dfrac52\right\}$ is a geometric sequence.

(3) Let $P_n$ be the product of all terms after the $n$-th product-growth operation. Then $b_n=\log_4P_n$. After the $(n+1)$-th product-growth operation, the product of newly inserted terms is $\dfrac{P_n^2}{4}$, thus $$P_{n+1}=P_n\cdot\frac{P_n^2}{4}=\frac{P_n^3}{4},\qquad b_{n+1}=3b_n-1.$$ So $$b_{n+1}-\frac12=3\left(b_n-\frac12\right),\quad b_1=2,$$ which gives $b_n=\dfrac12\cdot3^n+\dfrac12$.

After one sum-growth operation starting from $1,4$, the sequence becomes $1,5,4$, so $a_1=1+5+4=10$. From part (2), together with $a_1=10$, we get $a_n=\dfrac52\cdot3^n+\dfrac52$. Therefore $$n(a_n+b_n)=n(3^{n+1}+3).$$ Let $T_n=\sum_{k=1}^n k\cdot3^{k+1}$. Using the shifted-subtraction method: $$T_n = 1\cdot3^2+2\cdot3^3+\cdots+n\cdot3^{n+1},$$ $$3T_n = 1\cdot3^3+2\cdot3^4+\cdots+n\cdot3^{n+2}.$$ Subtracting: $$-2T_n = 3^2+3^3+\cdots+3^{n+1}-n\cdot3^{n+2} =\frac{9(3^n-1)}{2}-n\cdot3^{n+2}.$$ Hence $$T_n=\frac{9(3^n-1)}{4}+\frac{n\cdot3^{n+2}}{2}=\frac{(2n-1)\cdot3^{n+2}}{4}+\frac9{4}.$$ Hence $$\sum_{k=1}^n k(a_k+b_k)=T_n+3\sum_{k=1}^n k =\left(\frac{2n-1}{4}\right)3^{n+2}+\frac{3n(n+1)}2+\frac94.$$

(1) The number of inserted terms after $n$ growth operations is $k=2^n-1$, so for $n=4$, $k=15$.

(2) The sequence $\left\{a_n-\dfrac52\right\}$ is geometric because it satisfies $a_{n+1}-\frac52=3\left(a_n-\frac52\right)$, with common ratio $3$.

(3) Using the explicit forms of $a_n$ and $b_n$, the required sum is $$\sum_{k=1}^n k(a_k+b_k)=\left(\frac{2n-1}{4}\right)3^{n+2}+\frac{3n(n+1)}2+\frac94.$$ This comes from decomposing the sum into $\sum k3^{k+1}$ and $\sum k$, then simplifying.