Kenneth Rosen Edition 6th Exercise 6.1 Question 9d (Page No. 401)

Question

Solve the recurrence relation $a_n = a_{n-1} + 2n + 3, a_0 = 4$

Answer 1

The recurrence relation is n^2 + 4n + 4 

Comments

Thanks :)
 I had made the mistake of adding up all the constant terms and then looking for some pattern in them :p

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Welcome.:)

Answer 2

$a_{n}=a_{n−1}+2n+3$

$a_{0}=4$

$a_{n} - a_{n−1} = 2n+3$

$a_{1} - a_{0} = 2*1+3$

$a_{2} - a_{1} = 2*2+3$

$a_{3} - a_{2} = 2*3+3$

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$a_{n} - a_{n−1} = 2*n+3$

on adding we get

$a_{n} - a_{0} = 2*(1 + 2 +3 + 4 + 5 + 6 + .................+ n)$ + $(3+3+3+.....................+n-times)$

$a_{n} - a_{0} = (2*n*(n+1) )/2$ + $3*n$

$a_{n} - a_{0} = n*(n+1) $ + $3*n$

$a_{n} - a_{0} = n^2 + n + 3*n$

$a_{n} - a_{0} = n^2 + 4*n$

$a_{n}  = n^2 + 4*n + a_{0}$

$a_{n}  = n^2 + 4*n + 4$

Answer 3

$\large a_n = \left\{\begin{matrix} a_{n-1}+2n+3, n \ge 0\\ 4, n = 0 \end{matrix}\right. \\ \\ \Rightarrow \sum_{1 \le k \le n} a_k-a_{k-1} = \sum_{1 \le k \le n}2k+3 \\ \Rightarrow a_n - a_0 = 3n+ 2\sum_{1 \le k \le n} k = 3n + 2\binom{n+1}{2}\\ \Rightarrow a_n = 4 + 3n+2\binom{n+1}{2}$