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.