an+a(n+2)=5/2a(n+ 1)

a(n+2)-2a(n+ 1)= 1/2a(n+ 1-an = 1/2[a(n+ 1)-2an]

Therefore, [a (n+2)-2a (n+1)]/[a (n+1)-2an] = {1/2 [a (n+1)-2an]}

That is, {a(n+ 1)-2an} is a geometric series, and the common ratio is = 1/2.

2[an+a(n+2)]=5a(n+ 1)

an+a(n+2)=5/2a(n+ 1)

a(n+2)- 1/2a(n+ 1)= 2a(n+ 1)-an = 2[a(n+ 1)- 1/2an]

Therefore, [a (n+2)-1/2a (n+1)]/[a (n+1)-12an] = {2 [a (n+/kloc-0]

That is, {a(n+ 1)- 1/2an} is a geometric series, and the common ratio is 2.

2){a(n+ 1)-2an} is a geometric series with the ratio = 1/2.

a2-2a 1=2-2*5=-8

a(n+ 1)-2an=-8*( 1/2)^(n- 1)=-( 1/2)^(n-4)

a(n+ 1)-2an=-( 1/2)^(n-4)

Divide both sides of the equation by 2 (n+ 1).

a(n+ 1)/2^(n+ 1)-an/2^n==-8*( 1/4)^n

Cumulative:

a(n+ 1)/2^(n+ 1)-a 1/2=-8[ 1/4+( 1/4)^2+...+( 1/4)^n]=-8/3[ 1-( 1/4)^n]

a(n+ 1)/2^(n+ 1)-a 1/2=-8/3[ 1-( 1/4)^n]

a(n+ 1)/2^(n+ 1)=- 1/6+8/3( 1/4)^n

a(n+ 1)=2^(n+ 1)[- 1/6+8/3( 1/4)^n]

an= 1/6*[2^(6-n)-2^n]

tn=2^(n-3)*an= 1/6*[2^(6-n)-2^n]2^(n-3)= 1/6[8-( 1/8)*4^n]=4/3-( 1/48)*4^n

Tn=4/3-( 1/48)*4^n

Sn=4n/3- 1/48[4+4^2+....+4^n]=4n/3- 1/48*[-4/3( 1-4^n)/]=( 1-48n-4^n)/36

Sn= 1/36*( 1+48n-4^n)