Tính: 1 + 1/2 + 1/3 + ... + 1/62 + 1/63 + 1/64
Những câu hỏi liên quan
1+1/2+1/3+...+1/62+1/63+1/64
CMR: 1+1/2+1/3+1/62+1/63+1/64> 4
CMR:M=1+1/2+1/3+..+1/62+1/63+1/64>4
CMR 1+1/2+1/3+...+1/62+1/63+1/64 > 4
cmr:1+1/2+1/3+...+1/62+1/63+1/64>4
1, Tính :
a, A = 1.(-1)+3.(-1)^2+5.(-1)^3.7+(-1)^4+9.(-1)^5
b, 1-2-3+4+5-6-7+8+...+61-62-63-64
1, Tính :
a, A = 1.(-1)+3.(-1)^2+5.(-1)^3.7+(-1)^4+9.(-1)^5
b, 1-2-3+4+5-6-7+8+...+61-62-63-64
\(A=1.\left(-1\right)+3.\left(-1\right)^2+5.\left(-1\right)^3.7+\left(-1\right)^4+9.\left(-1\right)^5\)
\(A=1.\left(-1\right)+3.1+5.\left(-1\right).7+1+9.\left(-1\right)\)
\(A=\left(-1\right)+3+\left(-5\right).7+1+\left(-9\right)\)
\(A=-1+3-35+1-9\)
\(A=-41\)
Đúng 0
Bình luận (0)
CM : 1+ 1/2+1/3+1/4+1/5+.....+1/62+1/63+1/64 > 4
Chứng tỏ rằng 1+1/2+1/3+1/4+...+1/62+1/63+1/64>4
Ta có: A = 1/2+1/3+1/4+...+1/62+1/63+1/64
A = 1+(1/2+1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+1/10+...+1/16)+...+(1/17+1/18+....+1/32)+(1/33+1/34+...+1/64)
Ta có: 1/2+1/3+1/4>1/2+1/4+1/4=1
1/5+1/6+1/7+1/8>1/8+1/8+1/8+1/8=1/8.4=1/2
1/9 +1/10+...+1/16>1/16+1/16+...1/16=1/16.8=1/2
1/33+1/34+...+1/64>1/64+1/64+...+1/64=1/64.32=1/2
Vậy A > 4
Đúng 1
Bình luận (0)
Ta có A = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/64
A = 1 + (1/2 + 1/3 + 1/4) + (1/5 + 1/6 + ... + 1/8) + (1/9 + 1/10 + 1/11 + ... + 1/16) + (1/17 + 1/18 + 1/19 + ... + 1/32) + (1/33 + 1/34 + 1/35 + ... + 1/64)
=> A > 1 + (1/2 + 1/4.2) + 1/8.4 + 1/16.8 + 1/32.16 + 1/64.32
A > 1 + 1 + 1/2 + 1/2 + 1/2 + 1/2
A > 4 (DPCM).
Đúng 0
Bình luận (0)