If f(1)=3, and f(n+1)-f(n)=3(4^n-1), then for | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(1)=3$ and $f(n+1)-f(n)=3(4^n-1)$.We can write the recurrence as a telescoping sum:$f(n) = f(1) + \sum_{k=1}^{n-1} (f(k+1)-f(k))$$f(n) = 3

Vedclass · Accepted Answer

$4^n-3n+2$

Vedclass · Answer

(C) Given $f(1)=3$ and $f(n+1)-f(n)=3(4^n-1)$.We can write the recurrence as a telescoping sum:$f(n) = f(1) + \sum_{k=1}^{n-1} (f(k+1)-f(k))$$f(n) = 3 + \sum_{k=1}^{n-1} 3(4^k-1)$$f(n) = 3 + 3 \left( \sum_{k=1}^{n-1} 4^k - \sum_{k=1}^{n-1} 1 \right)$Using the geometric series sum formula $\sum_{k=1}^{n-1} 4^k = \frac{4(4^{n-1}-1)}{4-1} = \frac{4^n-4}{3}$:$f(n) = 3 + 3 \left( \frac{4^n-4}{3} - (n-1) \right)$$f(n) = 3 + (4^n-4) - 3(n-1)$$f(n) = 3 + 4^n - 4 - 3n + 3$$f(n) = 4^n - 3n + 2$

If $f(1)=3$, and $f(n+1)-f(n)=3(4^n-1)$, then for all $n \in N$, $f(n)=$

Similar Questions

What is the sum of the infinite series $1^2 + 2^2 x + 3^2 x^2 + \dots$?

The sum $11^3 + 12^3 + \dots + 20^3$ is:

If $b$ is the first term of an infinite $G.P.$ whose sum is $5$,then $b$ lies in the interval

For any integer $n \geq 1$,the sum $\sum_{k=1}^n k(k+2)$ is equal to

If $2.5+5.9+8.13+11.17+\ldots$ to $n$ terms $=an^3+bn^2+cn+d$,then $a-b+c-d=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module