(a)Given, A(n+1)= A(n) +2n+3, hence for n=5,

A(6)= A(5) +13

=A(4)+11+13 [ A(5)= A(4)+11]

= A(3)+9+11+13 [A(4)=A(3)+ 9]

= A(2)+7+9+11+13 [A(3)=A(2)+7]

=A(1)+5+7+9+11+13 [A(2)=A(1)+5]

=A(0)+3+5+7+9+11+13 [A(1)=A(0)+3]

= A(0)+48

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(b) Given,u(n+2)-2u(n+1)+u(n)=1, u(0)=2 and u(1)=-1

U(n+2)=2u(n+1)-u(n)+1, hence for n=4

U(6)= 2u(5)-u(4)+1

= 2[2u(4)-u(3)+1]-u(4)+1 {u(5)=2u(4)-u(3)+1}

=3u(4)-2u(3)+3

=3[2u(3)-u(2)+1]-2u(3)+3 {u(4)=2u(3)-u(2)+1}

=4u(3)-3u(2)+6

=4[2u(2)-u(1)+1]-3u(2)+6 {u(3)=2u(2)-u(1)+1}

=5u(2)-4u(1)+10

=5[2u(1)-u(0)+1]-4u(1)+10 {u(2)=2u(1)-u(0)+1}

=6u(1)-5u(0)+15

=-6-10+15

=-1

*****