How to solve given Recurrence Equation $a_{n} = 6n^2 + 2n + a_{n - 1}$?

Question

How to solve given Recurrence Equation $a_n = 6n^2 + 2n + a_{n - 1}$ ?

Given $a_{1} = 8$

Find the value of $a_{99}$ ?

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Is there any procedure or approach for such questions ?

Answer 1

$$a(n)=6\sum_{r=1}^nr^2+2\sum_{r=1}^nr+a(0)$$

Answer 2

This isn't "really" a recursive relation, since all you are doing is adding some terms to the previous value.

In other words, we can recover a non-closed formula for this quite easily:

$f(n)=\sum\limits_{i=1}^n 6i^2+2i+ c$. (where $c$ is a constant so that the formula holds for $f(1)$, in this case $c=0$, since $f(1)=6(1)^2+2(1)$ )

We can rewrite this as:

$f(n)=6\sum\limits_{i=1}^n i^2+2\sum\limits_{i=1}^n i=6\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}=n(n+1)(2n+2)=2n(n +1)^2$.

Note that I used the formula for the sum of consecutive squares and the formula for the sum of an arithmetic series.

Answer 3

$a_n=6n^2+2n+a_{n−1}$

$=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2} $

$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1$

$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+6.1^2+2.1$

$=6(n^2+(n−1)^2+...+2^2+1^2)+2(n+(n−1)+...+2+1)$

$=6\times \frac{n(n+1)(2n+1)}{6}+2\times \frac{n(n+1)}{2} $

$=n(n+1)(2n+1+1)$

$a_n=2n(n+1)^2$

for $n=99$, $a_{99}=2×99×(99+1)^2=198\times 10^4$. Hope it helps.