叠加与累乘

已知数列\(\left\{a_n\right\}\)满足\(a_1=\frac{1}{3},\frac{1}{a_{n+1}}=\frac{3}{a_{n}\cdot(a_{n}+3)}\),设\(b_n=\frac{1}{a_{n}+3},P_n=b_1\cdot{b_2}\cdot{b_3}\cdot...\cdot{b_n}\)
\(Q_n=b_1+b_2+...+b_n\),则\(3^{n+1}P_n+Q_n\)=____________.
分析：注意到\(P_n\)是一个数列的前\(n\)项积，因此考虑用累乘法求之；而\(Q_n\)是一个数列的前\(n\)项和，
所以考虑用列项想消法求之，有了以上分析，我们有下面的解法.
解：由\(\frac{1}{a_{n+1}}=\frac{1}{a_{n}\cdot(a_{n}+3)}\)得：\(b_n=\frac{a_n}{3a_{n+1}}=\frac{1}{a_n}-\frac{1}{a_{n+1}}\)
\(\therefore\) \(P_n=\frac{a_1}{3a_2}\cdot\frac{a_2}{3a_3}\cdot...\cdot\frac{a_n}{3a_{n+1}}=\frac{a_1}{3^{n}a_{n+1}}=\frac{1}{3^{n+1}a_{n+1}}\)
\(Q_n=\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}+...+\frac{1}{a_n}-\frac{1}{a_{n+1}}=\frac{1}{a_1}-\frac{1}{a_{n+1}}\)
\(\therefore\) \(3^{n+1}P_n+Q_n=\frac{1}{a_{n+1}}+\frac{1}{a_1}-\frac{1}{a_{n+1}}=\frac{1}{a_1}=3\)