S(n- 1)=2a(n- 1)-2 - ②
①-② an=2an-2a(n- 1)
∴an=2a(n- 1)
an/a(n- 1)=2
∴{an} is a geometric series with a common ratio of 2.
s 1=2a 1-2=a 1
∴a 1=2
∴an=a 1*q^(n- 1)=2*2^(n- 1)=2^n
(2)∫b(n+ 1)= bn+2
∴b(n+ 1)+2=bn+2+2
∴[b(n+ 1)+2]-[bn+2]=2
The sequence {bn+2} is a arithmetic progression with an error of 2.
b 1+2=3
∴bn+2=3+(n- 1)x2=3+2n-2=2n+ 1
∴bn=2n- 1
(3) The sequence {anbn} is the product of arithmetic progression and geometric progression. For the summation of this sequence, dislocation subtraction should be used.
dn=(2- 1)2^ 1+(4- 1)2^2+(6- 1)2^3+...+(2n- 1)2^n - ①
2dn=(2- 1)2^2+(4- 1)2^3+(6- 1)2^4+...+(2(n- 1)- 1)2^n+(2n- 1)2^(n+ 1)-②
①-②-DN = 2+2 [2 2+2 3+...+2 n]-2 (2n- 1) 2 n。
? =2+2[2(2^n- 1)-2]-2(2n- 1)2^n
? =4*2^n-6-2(2n- 1)2^n
∴Dn=2(2n- 1)2^n+6-4*2^n
=(4n-2-4)2^n+6
=(2n-3)2^(n+ 1)+6
∫[sin(nπ/2)]2 = 0, (n is an even number)
[sin (n π/2)] 2 = 1, (n is an odd number)
∫[cos(nπ/2)]2 = 1, (n is an even number)
[cos (n π/2)] 2 = 0, (n is an odd number)
∴T(2n)=a 1+a3+a5+..+a(2n- 1)-[b2+b4+b6+b8+...+b(2n)]
=2^ 1+2^3+2^5+...+2^(2n- 1)-[2*2- 1+2*4- 1+2*6- 1+...+2*2n- 1]
=[2( 1-4^n)]/( 1-4)-[2(2+4+6+8+..+2n)-n]
=[2(4^n- 1)]/3-2(n^2)-2n+n
=[2(4^n- 1)]/3-2(n^2)-n
You can ask if you don't understand.