If f(1)=3, and f(n+1)-f(n)=3(4n-1), then ∀ n ∈ N, f(n)=

Question

If f(1)=3, and f(n+1)-f(n)=3(4n-1), then ∀ n ∈ N, f(n)= (A) 4n-1 (B) 4n-5 n+4 (C) 4n-3 n+2 (D) 4n+4 n-5

Tardigrade Admin · Accepted Answer

We have, f(1)=3 f(n+1)-f(n)=3(4n-1) f(2)-f(1)=3(4-1) f(3)-f(2)=3(42-1) f(n)-f(n-1)=3(4n-1-1) Adding, we get f(n)-f(1)=3(4+42-4n-1)-3(n-1) f(n)-3=3((4(4n-1-1)/4-1))-3 n+3 f(n)-3=4n-4-3 n+3 f(n)=4n-3 n+2

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

$4^{n} - 1$

$4^{n} - 5 n + 4$

$4^{n} - 3 n + 2$

$4^{n} + 4 n - 5$

Q. If f(1)=3, and f(n+1)−f(n)=3(4n−1), then ∀ n∈N,f(n)=

4n−1

4n−5n+4

4n−3n+2

4n+4n−5

We have, f(1)=3 f(n+1)−f(n)=3(4n−1) f(2)−f(1)=3(4−1) f(3)−f(2)=3(42−1) f(n)−f(n−1)=3(4n−1−1) Adding, we get f(n)−f(1)=3(4+42−4n−1)−3(n−1) f(n)−3=3(4−14(4n−1−1)​)−3n+3 f(n)−3=4n−4−3n+3 f(n)=4n−3n+2

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

$4^{n} - 1$

$4^{n} - 5 n + 4$

$4^{n} - 3 n + 2$

$4^{n} + 4 n - 5$