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A = 1 + (1+ 1).2 + (1 + 2).3 + (1+3).4 + ...+ (1 + n-1). n
A = 1 + (2+1.2) + (3+ 2.3) + (4 + 3.4) + ....+ ( n + (n -1).n)
A = (1+ 2 + 3 + 4 + ...+ n) + (1.2 + 2.3 + 3.4 + .....+ (n-1).n)
Tính B = 1+ 2+ 3 + ...+ n = (n +1).n/ 2
C = 1.2+ 2.3 + 3.4 + ...+ (n-1).n
=> 3.C = 1.2.3 + 2.3.3 + 3.4.3 + ...+ (n-1).n.3
3C = 1.2.3 + 2.3. (4 -1) + 3.4.(5 - 2) + ... + (n -1).n [(n+ 1) - (n -2)]
3C = [1.2.3 + 2.3.4 + ....+ (n-1).n.(n +1)] - (1.2.3 + 2.3.4 + ... + (n-2)(n -1).n)
3C = (n -1).n (n +1) => C = (n -1).n.(n +1)/ 3
Vậy A = (n +1).n/ 2 + (n -1).n(n +1)/3
A=1-3+3^2-3^3+3^4-3^5+...+3^38-3^39+3^100
3A=3-3^2+3^3-3^4+3^5-3^6+...+3^99-3^100+3^101
3A+A=3-3^2+3^3-3^4+3^5-3^6+...+3^99-3^100+3^101+1-3+3^2-3^3+3^4-3^5+...+3^98-3^99+3^100
4A=3^101+1
A=(3^101+1)/4
2A = 2 + 1 + 1/2 + 1/2^2 + ... + 1/2^2011
A = 1 + 1/2 + .. + 1/2^2011 + 1/2^2012
2A - A = 2 + 1 + 1/2 + .. + 1/2^2011 - 1 - 1/2 - ... - 1/2^2011 - 1/2^2012
A = 2 - 1/2^2012
A = \(\frac{2^{2012}-2}{2^{2012}}\)
A = 1+1/2+1/2^2+1/2^3+.....+1/2^2012
2A= 2. (1+1/2+1/2^2+1/2^3+.....+1/2^2012)
2A= 2 + 1 + 1/2 + 1/2^2 + 1/2^3 + ...+ 1/2^2011
2A - A= (2 + 1 + 1/2 + 1/2^2 + 1/2^3+ ...+ 1/2^2011) - (1+1/2+1/2^2+1/2^3+.....+1/2^2012)
1A= 2 + 1 + 1/2 + 1/2^2 + 1/2^3 + ...+ 1/2^2011 - 1-1/2-1/2^2+1/2^3+.....+1/2^2012
1A= 2 - 1/2^2012
A= 2-1/2^2012
A= 2 - 1/2^2012
A = 1 + (1+ 1).2 + (1 + 2).3 + (1+3).4 + ...+ (1 + n-1). n
A = 1 + (2+1.2) + (3+ 2.3) + (4 + 3.4) + ....+ ( n + (n -1).n)
A = (1+ 2 + 3 + 4 + ...+ n) + (1.2 + 2.3 + 3.4 + .....+ (n-1).n)
Tính B = 1+ 2+ 3 + ...+ n = (n +1).n/ 2
C = 1.2+ 2.3 + 3.4 + ...+ (n-1).n
=> 3.C = 1.2.3 + 2.3.3 + 3.4.3 + ...+ (n-1).n.3
3C = 1.2.3 + 2.3. (4 -1) + 3.4.(5 - 2) + ... + (n -1).n [(n+ 1) - (n -2)]
3C = [1.2.3 + 2.3.4 + ....+ (n-1).n.(n +1)] - (1.2.3 + 2.3.4 + ... + (n-2)(n -1).n)
3C = (n -1).n (n +1) => C = (n -1).n.(n +1)/ 3
Vậy A = (n +1).n/ 2 + (n -1).n(n +1)/3
bài làm
A = 1 + (1+ 1).2 + (1 + 2).3 + (1+3).4 + ...+ (1 + n-1). n
A = 1 + (2+1.2) + (3+ 2.3) + (4 + 3.4) + ....+ ( n + (n -1).n)
A = (1+ 2 + 3 + 4 + ...+ n) + (1.2 + 2.3 + 3.4 + .....+ (n-1).n)
B = 1+ 2+ 3 + ...+ n = (n +1).n/ 2
C = 1.2+ 2.3 + 3.4 + ...+ (n-1).n
=> 3.C = 1.2.3 + 2.3.3 + 3.4.3 + ...+ (n-1).n.3
3C = 1.2.3 + 2.3. (4 -1) + 3.4.(5 - 2) + ... + (n -1).n [(n+ 1) - (n -2)]
3C = [1.2.3 + 2.3.4 + ....+ (n-1).n.(n +1)] - (1.2.3 + 2.3.4 + ... + (n-2)(n -1).n)
3C = (n -1).n (n +1)
=> C = (n -1).n.(n +1)/ 3
Vậy............
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