(3n–1)×2^2n-1,用错位相减法求总和
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let
S = 1.2^1+2.2^3+.+n.2^(2n-1) (1)
4S = 1.2^3+2.2^5+.+n.2^(2n+1) (2)
(2)-(1)
3S = n.2^(2n+1) - [2^1+2^3+...+2^(2n-1)]
= n.2^(2n+1) - (2/3)[2^(2n) -1 ]
an =(3n–1)×2^(2n-1)
= 3[n.2^(2n-1)] - 2^(2n-1)
Sn = a1+a2+...+an
=3S - (2/3)(2^(2n) -1)
=n.2^(2n+1) - (4/3)[2^(2n) -1 ]
= 4/3 + [2n- (4/3) ].2^(2n)
S = 1.2^1+2.2^3+.+n.2^(2n-1) (1)
4S = 1.2^3+2.2^5+.+n.2^(2n+1) (2)
(2)-(1)
3S = n.2^(2n+1) - [2^1+2^3+...+2^(2n-1)]
= n.2^(2n+1) - (2/3)[2^(2n) -1 ]
an =(3n–1)×2^(2n-1)
= 3[n.2^(2n-1)] - 2^(2n-1)
Sn = a1+a2+...+an
=3S - (2/3)(2^(2n) -1)
=n.2^(2n+1) - (4/3)[2^(2n) -1 ]
= 4/3 + [2n- (4/3) ].2^(2n)
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