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\(S=1+2+5+14+...+\frac{3^{n-1}+1}{2}\left(n\in N\right)\)
\(2S=2+4+10+28+...+\left(3^{n-1}+1\right)=S_1\)
\(2S=\left[1+1+1+...+n\right]+\left[1+3+9+...+3^{n-1}\right]\)
\(S_1=1+1+1+...+n=n\)
\(S_2=3+9+...+3^n\)
\(3S_2-S_2=2S_2=3^n-1\Rightarrow S_2=\frac{3^n-1}{2}\)
\(S=\frac{S_1+S_2}{2}=\frac{n+\frac{3^n-1}{2}}{2}=\frac{3^n+2n-1}{4}\)
Có 1 = \(\frac{3^0+1}{2}\)
2 = \(\frac{3^1+1}{2}\)
5 = \(\frac{3^2+1}{2}\)
14 = \(\frac{3^3+1}{2}\)
.......
=> S = \(\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+\frac{3^3+1}{2}+...+\frac{3^{n-1}+1}{2}\)
S = \(\frac{\left(3^0+3^1+3^2+3^3+...+3^{n-1}\right)+\left(1+1+1+1+...+1\right)}{2}\)
S = \(\frac{\left(3^0+3^1+3^2+3^3+...+3^{n-1}\right)+1.n}{2}\)
S = \(\frac{\left(3^0+3^1+3^2+3^3+...+3^{n-1}\right)+n}{2}\)
Đặt A = 30 + 31 + 32 + 33 +....+ 3n-1
=> 3A = 31 + 32 + 33 +....+ 3n
=> 2A = 3A - A = 3n - 30
=> A = \(\frac{3^n-1}{2}\)
Thay A vào S, ta có:
S = \(\frac{\frac{3^n-1}{2}+n}{2}\)
=> S = \(\frac{3^n-1}{4}+\frac{n}{2}\)
Hồ Thu Giang à, trong 4 đáp án ở bài Cóc vàng tài ba đó ko có cái này !
Đặt P=31-1+32-1+33-1+34-1+...+3n-1
=>P=30+31+32+33+...+3n-1
=>3.P=31+32+33+34+...+3n
=>3.P-P=31+32+33+34+...+3n-30-31-32-33-...-3n-1
=>2.P=3n-30
=>2.P=3n-1
=>\(P=\frac{3^n-1}{2}\)
Lại có: S=1+2+5+14+...+\(\frac{3^{n-1}+1}{2}\)
=>\(S=\frac{3^{1-1}+1}{2}+\frac{3^{2-1}+1}{2}+\frac{3^{3-1}+1}{2}+\frac{3^{4-1}+1}{2}+...+\frac{3^{n-1}+1}{2}\)
=>\(S=\frac{3^{1-1}+1+3^{2-1}+1+3^{3-1}+1+3^{4-1}+1+...+3^{n-1}+1}{2}\)
=>\(S=\frac{\left(3^{1-1}+3^{2-1}+3^{3-1}+3^{4-1}+...+3^{n-1}\right)+\left(1+1+1+1+...+1\right)}{2}\)
=>\(S=\frac{P+1.n}{2}\)
=>\(S=\frac{\frac{3^n-1}{2}+n}{2}\)
=>\(S=\frac{\frac{3^n-1}{2}+\frac{2n}{2}}{2}\)
=>\(S=\frac{\frac{3^n-1+2n}{2}}{2}\)
=>\(S=\frac{3^n-1+2n}{4}\)
c, \(\frac{-32}{-2^n}=4\)
\(\Rightarrow-2^n=-32:4\)
\(\Rightarrow-2^n=-8\)
\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)
d, \(\frac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\Rightarrow n=2\)
e, \(\frac{25^3}{5^n}=25\)
\(\Rightarrow5^n=25^3:25\)
\(\Rightarrow5^n=25^2\)
\(\Rightarrow5^n=5^4\Rightarrow n=4\)
i , \(8^{10}:2^n=4^5\)
\(\Rightarrow2^n=8^{10}:4^5\)
\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)
\(\Rightarrow2^n=2^{30}:2^{10}\)
\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)
k, \(2^n.81^4=27^{10}\)
\(\Rightarrow2^n=27^{10}:81^4\)
\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)
\(\Rightarrow2^n=3^{30}:3^{16}\)
\(\Rightarrow2^n=3^{14}\)
\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn
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+...+3n-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+...+3n-1 mình tính theo công thức nên tắt)