Answer:
(A)
[tex]C_n=3+2n[/tex]
(B)
[tex]C_1_0=23[/tex]
Step-by-step explanation:
(A)
we are given
[tex]C_n_+_1=C_n+2[/tex]
[tex]C_1=5[/tex]
Firstly, we will find few terms
[tex]C_2=C_1+2[/tex]
[tex]C_2=5+2[/tex]
[tex]C_2=7[/tex]
[tex]C_3=C_2+2[/tex]
[tex]C_3=7+2[/tex]
[tex]C_3=9[/tex]
[tex]C_4=C_3+2[/tex]
[tex]C_4=9+2[/tex]
[tex]C_4=11[/tex]
so, we will get terms as
5, 7 , 9 , 11
we can see that this is arithematic sequence
First term =5
common difference =d=7-5=2
now, we can use nth term formula
[tex]C_n=C_1+(n-1)d[/tex]
now, we can plug values
[tex]C_n=5+2(n-1)[/tex]
[tex]C_n=5+2n-2[/tex]
[tex]C_n=3+2n[/tex]
(B)
we can plug n=10
[tex]C_1_0=3+2\times 10[/tex]
[tex]C_1_0=23[/tex]