For f(n) = 10^n + 12(4^n+1) + 5, the difference f(n+1) - f(n) simplifies to which expression?

For $f(n) = 10^n + 12\left(4^{n+1}\right) + 5$ , the difference $f(n+1) - f(n)$ simplifies to which expression?

10^n + 36\left(4^{n+1}\right)

9\left(10^n + 4^{n+2}\right)

9\left(10^n + 4^{n+1}\right)

9\times10^n + 12\left(4^{n+1}\right)

Explanation

The constant 5 cancels, leaving

f(n+1)-f(n)=\left(10^{n+1}-10^n\right)+12\left(4^{n+2}-4^{n+1}\right)=10^n(10-1)+12\times4^{n+1}(4-1)=9\times10^n+36\times4^{n+1}

. Since

36\times4^{n+1}=9\times4\times4^{n+1}=9\times4^{n+2}

, the difference is

9\left(10^n+4^{n+2}\right)

, a multiple of 9. So if

f(n)

is divisible by 9 then so is

f(n+1)

Derived from ZIMSEC Further Mathematics 9187/1, November 2010, Q6

Danho has every sitting for this paper, with your progress tracked question by question, offline.