Question

Find the general term of {a_n}


`\displaystyle \large{a_1 = 4, \: a_(n + 1) = 2a_n + 1}`
Please use the form below to solve.

`\displaystyle \large{a_(n + 1) - x = p(a_n - x)}`
Show your work too. Thanks!​

Answer 1

From the given recurrence, it follows that

`a_(n+1) = 2a_n + 1 \\\\ a_(n+1) = 2(2a_(n-1) + 1) = 2^2a_(n-1) + 1 + 2 \\\\ a_(n+1) = 2^2(2a_(n-2)+1) + 1 + 2 = 2^3a_(n-2) + 1 + 2 + 2^2 \\\\ a_(n+1) = 2^3(2a_(n-3) + 1) + 1 + 2 + 2^2 = 2^4a_(n-3) + 1 + 2 + 2^2 + 2^3`

and so on down to the first term,

`a_(n+1) = 2^na_1 + \displaystyle \sum_(k=0)^(n-1)2^k`

(Notice how the exponent on the 2 and the subscript of a in the first term add up to n + 1.)

Denote the remaining sum by S ; then

`S = 1 + 2 + 2^2 + \cdots + 2^(n-1)`

Multiply both sides by 2 :

`2S = 2 + 2^2 + 2^3 + \cdots + 2^n`

Subtract 2S from S to get

`S - 2S = 1 - 2^n \implies S = 2^n - 1`

So, we end up with

`a_(n+1) = 4\cdot2^n + S \\\\ a_(n+1) = 2^2\cdot2^n + 2^n-1 \\\\ a_(n+1) = 2^(n+2) + 2^n - 1 \\\\\implies \boxed{a_n = 2^(n+1) + 2^(n-1) - 1}`