Question

If $f(1)=3$, and $f(n+1)-f(n)=3\left(4^{n}-1\right)$, then $\forall$ $n \in N, f(n)=$

• A. $4^{n}-1$
• B. $4^{n}-5 n+4$
• C. $4^{n}-3 n+2$
• D. $4^{n}+4 n-5$

Answer 1

Answer: (C)

We have, $f(1)=3$
$f(n+1)-f(n)=3\left(4^{n}-1\right) $
$f(2)-f(1)=3(4-1)$
$f(3)-f(2)=3\left(4^{2}-1\right) $
$f(n)-f(n-1)=3\left(4^{n-1}-1\right)$
Adding, we get
$f(n)-f(1)=3\left(4+4^{2}-4^{n-1}\right)-3(n-1)$
$f(n)-3=3\left(\frac{4\left(4^{n-1}-1\right)}{4-1}\right)-3 n+3$
$f(n)-3=4^{n}-4-3 n+3$
$f(n)=4^{n}-3 n+2$