áp dụng quy tắc 

số số hạng= (số cuối-số đầu) chí cho khoảng cách rồi cộng với 1

Tổng=(số đầu +số cuối ) nhân với số số số hạng rồi chia cho 2

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

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

2S=(3⁰+3¹+...+3^n-1)+(1+1+...+1)                     (n số hạng 1)

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

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

(chỗ 3⁰+3¹+...+3^n-1 mình tính theo công thức nên tắt)

S=(3^0+1/2)+(3^1/2+1/2)+(3^2/2+1/2)+....+(3^n-1/2+1/2)

=n*1/2+1/2*(3^0+3^1+3^2+...+3^n-1)

=n^2/2+(3^n-1/4)=3^n+2-1/4

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

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

\(=\frac{\left(3^0+1\right)+\left(3^1+1\right)+\left(3^2+1\right)+.....+\left(3^{x-1}+1\right)}{2}\)

\(=\frac{\left(1+3+3^2+.....+3^{x-1}\right)+x}{2}\)

Đặt \(A=1+3+3^2+....+3^{x-1}\)

\(3A-A=\left(3+3^2+....+3^x\right)-\left(1+3+....+3^{x-1}\right)\)

\(2A=3^x-1\Rightarrow A=\frac{3^x-1}{2}\)

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

Đặt P=3^1-1+3^2-1+3^3-1+3^4-1+...+3^n-1

=>P=3⁰+3¹+3²+3³+...+3^n-1

=>3.P=3¹+3²+3³+3⁴+...+3ⁿ

=>3.P-P=3¹+3²+3³+3⁴+...+3ⁿ-3⁰-3¹-3²-3³-...-3^n-1

=>2.P=3ⁿ-3⁰

=>2.P=3ⁿ-1

=>\(P=\frac{3^n-1}{2}\)

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

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

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

=>\(S=\frac{\left(3^{1-1}+3^{2-1}+3^{3-1}+3^{4-1}+...+3^{n-1}\right)+\left(1+1+1+1+...+1\right)}{2}\)

=>\(S=\frac{P+1.n}{2}\)

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

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

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

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