Cho b=1+2+2mu2+2mu3+...+2mu6,A=2mu2+2mu3+2mu4+..+2mutam chứng minh rằng A=4B
Chung to tong sau co chia het cho 3 hay k
2+2mu2+2mu3+2mu4+2mu5+2mu6+2mu7+2mu8+2mu9+2mu10
Chứng tỏ rằng A=2+2mu2+2mu3+2mu4+chấm chấm chấm +2100 chia hết cho 6
A=4+2mu2+2mu3+2mu4+...+2mu20
A= 4+2.2+2.2.2+2.2.2.2+.......+{2.2.2.2.2.....} có 20 thừa số 2
Có số số hạng ở trong khoảng số 2 là:
(20-2)+1=19(số)
Có 20 thừa số 2 suy ra:20.2=40
Tổng là:
(40+2)*19:2=399
A=4+399
A=403
**** nhé Hương Linh xinh xắn
A=2+2mu2+2mu3+2mu4+2mu5+.....+2mu200
a, thu gọn A
b, tìm chữ số tận cùng của A
làm ơn giúp mình với!!
a. \(A=2+2^2+2^3+2^4+2^5+...+2^{200}\)
\(2A=2^2+2^3+2^4+2^5+2^6+...+2^{201}\)
\(2A-A=2^{201}-2\)
b. \(Tacó:2^1=2;2^2=4;2^3=8;2^4=..6;2^5=2;2^6=4;...\)
Cứ dựa theo quy luật trên, ta có:
\(2^{4k+1}=..2;2^{4k+2}=...4;2^{4k+3}=...8;2^{4k}=..6\)
\(=>2^{201}=2^{4.50+1}\)
\(=>2^{101}=...2\)
\(=>2^{201}-2=..2-2\)\(=0\)
Vậy A có tân cùng là 0
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 \(A+\frac{1}{2^{10}}=\frac{2}{2^{10}}-1+\frac{2}{10}=1\)
1.tính
a=1+2mu2+2mu4+2mu6.... +2mu96+2mu98+2mu100
Ta có; \(A=1+2^2+2^4+...+2^{100}\)
\(\Rightarrow4A=2^2A=2^2+2^4+...+2^{102}\)
\(\Rightarrow3A=4A-A=\left(2^2+...+2^{102}\right)-\left(1+...+2^{100}\right)\)
\(\Rightarrow3A=2^{102}-1\Rightarrow A=\frac{2^{102}-1}{3}\)
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
tim a biet a=2+2mu2+2mu3+...+2mu60
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a = 2 + 22 + 23 + ... + 260
2a = 2 . 2 + 22 . 2 + 23 . 2 + ... + 260 . 2
2a = 22 + 23 + 24 + ... + 261
2a - a = 261 - 2
a = 261 - 2
Vậy : a = 261 - 2