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已知数列an的前n项和

已知数列an的前n项和

(1)Sn=(n^2+n)/散碧2n=1, a1=1an = Sn - S(n-1) = (1/裂喊2)( 2n-1+1) = n=> an = n(2)bn=2^(an)+(-1)^n.an =2^n +(-1)^n.niebn= 2^n - n if n is odd = 2^n +n if n is evenS(2n) =[ b1+b3+...+b(2n-1) ]+ [b2+b4+...+b(2n) ]=(2/冲源举3)( 2^(2n) -1 ) - n^2 + (4/3)(2^(2n) -1) + (n+1)n=2( 2^(2n) -1 ) +n