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Five compulsory questions in senior high school mathematics
Series {a? n? } met one? = 1,a? n? =3a? n- 1? -4n+6(n≥2,n∈N*)。

(1) let b? n? =a? n? -2n, verification: sequence {b? n? } is a geometric series;

(2) Find the sequence {a? n? {and the first n items of s? n? .

Solution: b? n? =a? n? -2n=3a? n- 1? -4n+6-2n=3a? n- 1? -6n+6=3a? n- 1? -6(n- 1)=3[a? n- 1? -2(n- 1)]=3b? n- 1?

So b? n? /b? n- 1? =3= constant, ∴{b? n? } is the first item b? =a? -2= 1-2=- 1, geometric series of q=3: b? n? =-3? ;

Answer? n? =b? n? +2n, the sum of the first n terms: s? n? =∑b? n? +∑2n=-(3? - 1)/2+(2+4+6+.....+2n)

=( 1-3? )/2+(2+2n)n/2=( 1-3? )/2+n(n+ 1)

Note: ∑b? n? Is the sum of the first n terms of geometric series; ∑2n is the sum of the first n terms of arithmetic progression, the first term is 2, and the tolerance is 2.