Let a sequence be defined $a_1 = 0$ and $a_{n+1} - a_n=4n+3$ for all $n \ge1$ and $n \in N$.
- Find $a_k$ in terms of $k$.
My approach:-
Since common difference is a linear polynomial, I assumed the $k^{th}$ term to be a quadratic polynomial like $ak^2 + bk+c$.
Now,
$a_{k+1} - a_k=4k+3$
$a(k+1)^2 + b(k+1)+c-ak^2 - bk-c=4k+3$
$2ak + a+b=4k+3$
By comparing coefficients, $ a=2$ and $b=1$
For $c$, we put the value of $a_1$ and get $c=-3$
Hence, $a_k=2k^2+k-3$ or $a_k=(2k+3)(k-1)$
Is there any other method to solve this problem wherein I don't have to assume a polynomial and then find the general term.
- Find:- $$\lim_{n \to \infty} \frac{\sqrt{a_{4^0n}}+\sqrt{a_{4^1n}}+...+\sqrt{a_{4^{10}n}}}{\sqrt{a_{2^0n}}+\sqrt{a_{2^1n}}+...+\sqrt{a_{2^{10}n}}}$$
I tried to convert it into limit as a sum form but it isn't fruitful. Need hints.
For 1.
The equation is a linear recurrence of the first order. Such equations are known to hae a solution wich is the sum of the general solution of the homogeneous equation (RHS=0) and a particular solution of the inhomogeneous equation.
Then
$$a_{n+1}-a_n=0$$ obviously has the general solution $$a_n=C$$ for some constant,
and the first order difference of a polynomial is a polynomial of degree one less and leading coefficient equal to the degree, so we can try
$$a_n=2n^2+C'n$$
and
$$2(n+1)^2+C'(n+1)-2n^2-C'n=4n+2+C'$$ giving $C'=1$.
Hence
$$a_n=2n^2+n+C$$ and there are no other solutions. With $a_1=0$, you indeed select $$C=-3.$$