how to find $a_{50}$ from a recursive term Given $a_{n+1}=a_n+2n+3,a_1=3$
How can I find $a_{50}$?
I can compute $a_2,a_3,...,a_{50}$
But it's a long way. Is there any smart technique to compute?
Thanks.
 A: Do a "list":
$$\begin{align}&a_1=3\\&a_2=a_1+2+3\\&\ldots\\&a_n=a_{n-1}+2(n-1)+3\\&a_{n+1}=a_n+2n+3\end{align}$$
Now sum up both sides
$$\sum_{k=1}^{n+1}a_k=\sum_{k=1}^na_k+3(n+1)+2\sum_{k=1}^nk\implies$$
$$a_{n+1}=3(n+1)+2\sum_{k=1}^nk=3(n+1)+n(n+1)=(n+3)(n+1)$$
A: Hint: What are the values for the first eg. five values of $n$? What can you tell from those values?
A: Compute the first few values of $a_n$:
$$3,\ 8,\ 15,\ 24,\ 35,\ 48\dots$$
Note that between two consecutive values the difference is always an odd number. The difference starts from $5$ (which is given by $8 - 3$).
The sum of all the odd numbers from $1$ to $2n - 1$ is $n^2$. So the sum from $7$ to $2n - 1$ is $n^2 - 4$. Note that we have offset the sequence by one element and we have to start from $3$, so an alternative representation for $a_n$ is
$$a_n = (n + 1)^2 - 1.$$
We conclude that
$$a_{50} = 51^2 - 1 = 2600.$$
A: Other then methods above here is a general method which can work in many cases characteristic equation,but first you have to transform $a_{n+1}$ as combination of $a_k$'s without the constant terms and $n$ terms
$$a_{n+1}=a_n+2n+3\\a_n=a_{n-1}+2n+1\text{ I plugged n=n-1 into first equation}\\a_{n+1}-a_n=a_n-a_{n-1}+2\\a_{n+1}=2a_n-a_{n-1}+2\\a_n=2a_{n-1}-a_{n-2}+2\\a_{n+1}-a_n=2a_n-3a_{n-1}+a_{n-2}\\a_{n+1}-3a_n+3a_{n-1}-a_{n-2}=0$$
Now plug in $a_{n+1}=t^{n+1}$
$$t^{n+1}-3t^n+3t^{n-1}-t^{n-2}=0\\t^3-3t^2+3t-1=0\\(t-1)^3=0$$
Now since the roots of the polynomial are $t_1=t_2=t_3=1$
We have that $$a_n=k_1t_1^n+nk_2t_1^n+n^2k_3t_1^n=k_1+nk_2+n^2k_3$$
Now plugging in $n=1,2,3$ you'll get the coefficients
$$a_1=k_1+k_2+k_3=3\\a_2=k_1+2k_2+4k_3=8\\a_3=k_1+3k_2+9k_3=15\\k_2+3k_3=5\\2k_2+8k_3=12\\2k_2+6k_3=10\\k_3=1\\k_2=2\\k_1=0\\a_n=1n^2+2n+0=n(n+2)$$
