Question

两道高中数学题!请用纸写出过程，不会答不要乱答，会停止搜答案!

Answer 1

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Answer 2

#2
(1)
2Sn= 3^n +4
n=1, a1= 7/2
for n>=2
an = Sn -S(n-1)
2an= 3^n - 3^(n-1)
an = 3^(n-1)
ie
an
=7/2 ; n=1
=3^(n-1) ; n=2,3,4,...
S1 = a1 =7/2
for n>=2
Sn = 7/2 + 3^1+3^2+...+3^(n-1)
= 7/2 + (3/2)(3^(n-1) - 1)
(2)
bn=3n.an
Tn= b1+b2+...+bn
n=1, T1 = 3a1 = 21/2
for n>=2
Tn
= 21/2 + 3[ 2.3^1+3.3^2+....+n.3^(n-1) ]
=21/2 + 3S
=21/2 -3[3/4 + (1/4)(2n-1).3^n]
=33/4 + (3/4)(2n-1).3^n
consider
S =2.3^1+3.3^2+....+n.3^(n-1) (1)
3S = 2.3^2+3.3^3+....+n.3^n (2)
(2)-(1)
2S = n.3^n - [ 3^2+3^3+...+3^(n-1) ] - 6
2S = n.3^n - (9/2)[ 3^(n-2) -1 ] - 6
2S = n.3^n - (1/2).3^n - 3/2
S = -3/4 + (1/4)(2n-1).3^n
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####下一题
a1=2/3
a(n+1)= (1/2)an + 1/2
a(n+1) - 1 = (1/2)(an - 1)
=> {an - 1} 是等比数列, q=1/2
an - 1= (1/2)^(n-1) .(a1 - 1)
an - 1= -(1/3)(1/2)^n
an = 1-(1/3)(1/2)^n
Sn = a1+a2+...+an = n/2 -(1/3) [ 1- (1/2)^n ]