Question

Biết rằng s=1+2.3+\(3.3^2+...+11.3^{10}\)=a+\(\dfrac{21.3^b}{4}\), với a là số hữu tỉ, b là số nguyên. Tính \(P=a+\dfrac{b}{4}\)

Answer 1

\(S=1.3^0+2.3^1+3.3^2+...+11.3^{10}\)

\(3S=1.3^1+2.3^2+...+11.3^{11}\)

\(\Rightarrow S-3S=1+3^1+3^2+...+3^{10}-11.3^{11}\)

\(\Rightarrow-2S=1.\dfrac{3^{11}-1}{3-1}-11.3^{11}\)

\(\Rightarrow-2S=\dfrac{1}{2}.3^{11}-\dfrac{1}{2}-11.3^{11}\)

\(\Rightarrow-2S=-\dfrac{21.3^{11}+1}{2}\)

\(\Rightarrow S=\dfrac{1}{4}+\dfrac{21.3^{11}}{4}\)