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Review problems of mathematical sequence in senior three
Solution:

( 1)

2an = Sn+2 is the arithmetic average of Sn and 2.

When n= 1, 2a1= s1+2 = a1+2a1= 2.

When n=2, 2a2 = S2+2 = a1+a2+2a2 = a1+2 = 2+2 = 4.

(2)

When n= 1, a 1=2.

When n≥2, 2an = sn+22a (n-1) = s (n-1)+2.

2an-2a(n- 1)= Sn+2-S(n- 1)-2 = Sn-S(n- 1)= an

an=2a(n- 1)

An/a(n- 1)=2, which is a fixed value.

The sequence {an} is a geometric series with 2 as the first term and 2 as the common ratio. General formula an=2?

X=bn y=b(n+ 1) is substituted into the linear equation: b(n+ 1)=bn +2.

b(n+ 1)-bn=2

And b 1=2.

Sequence {bn} is a arithmetic progression, the first term is 2, and the tolerance is 2. The general formula bn=2n.

(3)

cn=anbn=2n×2?

Tn=a 1b 1+a2b2+...+anbn

=2( 1×2+2×2? +3×2? +...+n×2? )

2Tn=2[ 1×2? +2×2? +...+(n- 1)×2? +n×2^(n+ 1)]

Tn-2Tn=-Tn=2[2+2? +...+2? -n×2^(n+ 1)]

=2[2×(2? - 1)/(2- 1)-n×2^(n+ 1)]

=2[2^(n+ 1)-2 -n×2^(n+ 1)]

=2[( 1-n)×2^(n+2) -2]

=( 1-n)×2^(n+3) -4

Tn=(n- 1)×2^(n+3) +4 .

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