Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S=1+3+3^2+3^3+3^4+.....+3^{16}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+.....+\left(3^{2014}+3^{2015}+3^{2016}\right)\)
\(=1\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+......+3^{2014}\left(1+3+3^2\right)\)
\(=1.13+3^3.13+.....+3^{2014}.13\)
\(=13\left(1+3^3+....+3^{2014}\right)⋮13\)
\(\Rightarrow S⋮13\)
Ta có : S1 = 1 + (-3) + 5 + (-7) + .... + 17
= (1 - 3) + (5 - 7) + (9 - 11)+ (13 - 15) + 17
= -2 + -2 + -2 + -2 + 17
= -2 x 4 + 17
= -8 + 17
S1 = 9
S2 = (4 - 2) + (8 - 6) + (12 - 10) + (16 - 14) + -18
= 2 x 4 - 18
S2 = -10
S1 + S2 = 9 - 10 = -1
S1=1+(-3)+5+(-7)+...+17.
S1=-2+(-2)+....+(-2).(9 số -2).
S2=-2+4+(-6)+....+(-18)
S2=-2+(-2)+...+(-2).(9 số -2).
=> (-2).(9+9)=-36.
Ta có: S = 3+3/2+3/2^2+3/2^3+...+3/2^9
1/2.S = 3/2+3/2^2+3/2^3+3/2^4+...+3/2^10
\(\Rightarrow\) S-1/2.S = 3 - 3/2^10
\(\Rightarrow\) 1/2.S = 3 - 3/2^10
\(\Rightarrow\) S = (3 - 3/2^10) : 1/2
\(\Rightarrow\) S = 6 - 6/2^10
Nếu đúng thì cho mk biết nha
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}=3\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(=3\left(2-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^8}-\frac{1}{2^9}\right)=3\left(2-\frac{1}{2^9}\right)=6-\frac{3}{2^9}\)
\(S=3\left(1+\frac{1}{2^{ }}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(2S=3\left(\frac{2}{2^0}+\frac{2}{2^1}+\frac{2}{2^2}+...+\frac{2}{2^9}\right)=3\left(2+1+\frac{1}{2^{ }}+...+\frac{1}{2^8}\right)\)\(2S-S=S=3\left(2+1+\frac{1}{2^1}+...+\frac{1}{2^8}\right)-3\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)=3.\left(2-\frac{1}{2^9}\right)=3.\frac{2^{10}-1}{2^9}\)