S=2+2mu2+2mu3+....+2mu100;hay chung to rang S chia het cho 15 va chu so tan cung cua S
1\2+1\2mu2+1\2mu3+......+1\2mu100
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
tinh S=1/2+1/2mu2+1/2mu3+...+1/2mu20
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{20}}\)
=> \(2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\)
=> \(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
=> \(S=1-\frac{1}{2^{20}}\)
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
Cho S =1+2+2mu2+2mu3+...+2mu9+2mu10+2mu11.Hay so sanh Svoi 5*2mu10
tính:
2mu100-2mu99-....................-2mu2-2mu1
mu=mũ
ai đúng mik tĩk cho
tính:
2mu100-2mu99-....................-2mu2-2mu1
mu=mũ
ai đúng mik tĩk cho
\(A=2^{100}-2^{99}-...-2^2-2\)
\(2A=2^{101}-2^{100}-...-2^3-2^2\)
\(2A-A=2^{101}-2^{100}-...-2^3-2^2-2^{100}+2^{99}+...2^2+2\)
\(A=2^{101}-\left(2^{100}-2^{100}+2^{99}-2^{99}+...+2^2-2^2+-2\right)\)
\(A=2^{101}+2\)
chứng minh C= 2+2mu2+2mu5+...+2mu100 chia hết cho 31
tim a biet a=2+2mu2+2mu3+...+2mu60
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5454545454545454444445445445+5445455454*1541646569-156456
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
tính:
2mu100-2mu99-....................-2mu2-2mu1
mu=mũ
2100 -299=21
298-297=21
=> từ 21 -> 2100 có 50 số 21
=>2100-299-298-...-21=21.50=250