Question

       Tính tổng     \(S=1+2+5+14+...+\frac{3^{n-1}+1}{2}\)

Answer 1

ai giup mink nha

Answer 2

\(S=1+2+5+14+...+\frac{3^{n-1}+1}{2}\)

\(=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+\frac{3^3+1}{2}+...+\frac{3^{n-1}+1}{2}\)

\(=\frac{\left(3^0+3^1+3^2+3^3+...+3^{n-1}\right)+\left(1+1+1+1+...+1\right)}{2}\)(tổng thứ 2 trên tử có n chữ số 1)

Đặt \(K=3^0+3^1+3^2+3^3+...+3^{n-1}\)

\(\Rightarrow3K=3^1+3^2+3^3+3^4+...+3^n\)

\(\Rightarrow3K-K=3^1+3^2+3^3+3^4+...+3^n\)\(-3^0-3^1-3^2-3^3-...-3^{n-1}\)

\(\Rightarrow2K=3^n-1\Rightarrow K=\frac{3^n-1}{2}\)

\(\Rightarrow S=\frac{\frac{3^n-1}{2}+n}{2}=\frac{3^n+2n-1}{4}\)

Vậy \(S=\frac{3^n+2n-1}{4}\)

Answer 3

mk cx nghĩ vậy nhưng hỏi cho chắc

Answer 4

Quy luật:

\(S=\frac{3^{1-1}+1}{2}+\frac{3^{2-1}+1}{2}+\frac{3^{3-1}+1}{2}+...+\frac{3^{n-1}+1}{2}\)

\(S=\frac{\left(3^{1-1}+3^{2-1}+3^{3-1}+...+3^{n-1}\right)+\left(1+1+1+...+1\right)}{2}\) có n chữ số 1

\(S=\frac{\frac{3^n-1}{2}+n}{2}=3^n-1+\frac{n}{2}\)