1\2+1\2mu2+1\2mu3+......+1\2mu100
H24
Những câu hỏi liên quan
cho biểu thuc B =2+2mu2+2mu3+.......+2mu99+2mu100
A. Chung minh rang Bchia het cho 31
b.tim x de 2mu2x-1-1=B
B= ( 2 + 2^2 + 2^3 + 2^4 + 2^5) + 2^5. ( 2 + 2^2 + 2^3 + 2^4 + 2^5)+....+ 2^95 ( 2 + 2^2 + 2^3 + 2^4 + 2^5)
= 62.(1 + 2^5 + ... + 2^95 ) chia hết cho 62
Suy ra B chia hết cho 31
Đúng 0
Bình luận (0)
S=2+2mu2+2mu3+....+2mu100;hay chung to rang S chia het cho 15 va chu so tan cung cua S
Cho b=1+2+2mu2+2mu3+...+2mu6,A=2mu2+2mu3+2mu4+..+2mutam chứng minh rằng A=4B
\(B=1+2+2^2+...+2^6.\)
\(=>4B=2^2+2^3+...+2^8\)\(\left(1\right)\)
\(A=2^2+2^3+...+2^8\)\(\left(2\right)\)
Từ (1) và (2)
=> A = 4B
Đúng 0
Bình luận (0)
tinh S=1/2+1/2mu2+1/2mu3+...+1/2mu20
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{20}}\)
=> \(2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\)
=> \(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
=> \(S=1-\frac{1}{2^{20}}\)
Đúng 0
Bình luận (0)
1.tính
a=1+2mu2+2mu4+2mu6.... +2mu96+2mu98+2mu100
Ta có; \(A=1+2^2+2^4+...+2^{100}\)
\(\Rightarrow4A=2^2A=2^2+2^4+...+2^{102}\)
\(\Rightarrow3A=4A-A=\left(2^2+...+2^{102}\right)-\left(1+...+2^{100}\right)\)
\(\Rightarrow3A=2^{102}-1\Rightarrow A=\frac{2^{102}-1}{3}\)
Đúng 0
Bình luận (0)
B = 1+2+2mu2+2mu3+...+2mu2008 phan 1-2mu2009
A=1/2+1/2mu2+1/2mu3+.....+1/2mu10
Chứng minh: A+1/2mu10=1
\(\Rightarrow\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\)
\(\Rightarrow\) \(\frac{1}{2}A=A-\frac{1}{2}=\frac{1}{2^{10}}-\frac{1}{2}\)
Vậy \(A=\left(\frac{1}{2^{10}}-\frac{1}{2}\right):\frac{1}{2}=\frac{2}{2^{10}}-1\)
Do đó \(A+\frac{1}{2^{10}}=\frac{2}{2^{10}}-1+\frac{2}{10}=1\)
Đúng 0
Bình luận (0)
Cho S =1+2+2mu2+2mu3+...+2mu9+2mu10+2mu11.Hay so sanh Svoi 5*2mu10
1+2+2mu2+2mu3+...+2mu2006
Nhanh lên nha
Mik đang gấp