Sequences & Series – Problem 5: Find the general formulas of and

Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence $\{a_n\}$, with $a_6=S_6=6$. A geometric sequence $\{b_n\}$ satisfies $b_1=a_4$, $b_2=a_5$.

(1) Find the general formulas of $\{a_n\}$ and $\{b_n\}$.

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

(1) From the conditions, $S_5=S_6-a_6=0$. Also, $S_5=\dfrac{5(a_1+a_5)}2=5a_3$, so $a_3=0$. Hence the common difference is $$d=\frac{a_6-a_3}{3}=2,$$ therefore $$a_n=a_3+(n-3)d=2n-6.$$

Also, $b_1=a_4=2$, $b_2=a_5=4$, so the common ratio is $q=2$, and $$b_n=2^n.$$

(2) $$\sum_{k=1}^n(a_k+b_k)=\sum_{k=1}^n(2k-6+2^k) =\sum_{k=1}^n(2k-6)+\sum_{k=1}^n2^k.$$ Compute: $$\sum_{k=1}^n(2k-6)=n^2-5n,\qquad \sum_{k=1}^n2^k=2^{n+1}-2.$$ So $$\sum_{k=1}^n(a_k+b_k)=n^2-5n-2+2^{n+1}.$$

(1) The arithmetic sequence is $a_n=2n-6$, and the geometric sequence is $b_n=2^n$. These follow from $a_6=S_6=6$, which gives $a_3=0$ and common difference $2$, together with $b_1=a_4$, $b_2=a_5$.

(2) The required partial sum is $$\sum_{k=1}^n(a_k+b_k)=n^2-5n-2+2^{n+1}.$$ It is obtained by adding the arithmetic-part sum $\sum(2k-6)$ and the geometric-part sum $\sum 2^k$.