Question

7) Given that Sn=3-2^(-4n)

Answer 1

Answer:

Step-by-step explanation:

$s_{n}=3-2^{-4n}\\s_{n+1}=3-2^{-4(n+1)}=3-2^{-4n} *2^{-4}\\s_{n+1}-s_{n}=3-2^{-4n}*2^-4-3+2^{-4n}$

$=2^{-4n}-2^{-4n}*2^{-4}\\=2^{-4n}(1-2^{-4})\\=2^{-4n}(\frac{2^4-1}{2^4})\\=2^{-4n}\frac{15}{16}\\U_{n+1}=15*2^{-4(n+1)}$