设Sn为数列an的前n项和,Sn=(-1)的n次幂-1/2的n次幂,n∈正整数,则(1)a3=(2)S1+S2+.
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设Sn为数列an的前n项和,Sn=(-1)的n次幂-1/2的n次幂,n∈正整数,则(1)a3=(2)S1+S2+.
s(n) = (-1)^n - 1/2^n,
a(1) = s(1) = (-1)^1 - 1/2 = -3/2,
a(3) = s(3) - s(2) = (-1)^3 - 1/2^3 - [(-1)^2 - 1/2^2] = -1 - 1/8 - [1 - 1/4] = -9/8 - 3/4 = -15/8
s(1)+s(2)+...+s(n) = [(-1)^1 + (-1)^2 + ...+ (-1)^n] - [1/2 + 1/2^2 + ...+ 1/2^n]
= (-1)[1 + (-1) + (-1)^2 + ...+ (-1)^(n-1)] - (1/2)[1 + (1/2) + ...+ (1/2)^(n-1)]
= -[1-(-1)^n]/(1+1) - (1/2)[1-(1/2)^n]/(1-1/2)
= -[1 - (-1)^n]/2 - [1 - (1/2)^n]
= -3/2 + (1/2)(-1)^n + (1/2)^n
a(1) = s(1) = (-1)^1 - 1/2 = -3/2,
a(3) = s(3) - s(2) = (-1)^3 - 1/2^3 - [(-1)^2 - 1/2^2] = -1 - 1/8 - [1 - 1/4] = -9/8 - 3/4 = -15/8
s(1)+s(2)+...+s(n) = [(-1)^1 + (-1)^2 + ...+ (-1)^n] - [1/2 + 1/2^2 + ...+ 1/2^n]
= (-1)[1 + (-1) + (-1)^2 + ...+ (-1)^(n-1)] - (1/2)[1 + (1/2) + ...+ (1/2)^(n-1)]
= -[1-(-1)^n]/(1+1) - (1/2)[1-(1/2)^n]/(1-1/2)
= -[1 - (-1)^n]/2 - [1 - (1/2)^n]
= -3/2 + (1/2)(-1)^n + (1/2)^n
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