finding explicit formula The question ask us to guess an explicit formula for the sequence
$$y_k = y_{k-1} + k^2 ,$$ for all integers $k$ greater than or equal to 2 and 
$y_1 = 1$
Can someone help me with this? 
so far what I got is this:
$y_2 = 1 + 2^2$
$y_3 = 1 + [2^2 + 3^2]$
$y_4 = 1 + [2^2 + 3^2 + 4^2]$
How do I convert those in the bracket into a formula?
Thanks in advance.
 A: This is a well known series, given by $y_k= 1/6\, k(k+1)(2k+1)$. There are several methods to prove it. One of the standard approaches is by induction. A simple and intuitive method is the following:
1 - write the sequence of differences between consecutive terms: $1, 4, 9, 16...$;
2 - write the sequence of differences between consecutive terms of the previous sequence: $1, 3, 5, 7...$;
3 - again, write the sequence of differences between consecutive terms of the previous sequence: $2, 2, 2, 2...$. 
Because the terms of this third sequence are  constant, the sum is a cubic of the form 
$ak^3+bk^2+ck+d$.
Since the sum is zero for $k=0$, it follows that $d=0$. Then,  we can identify coefficients rewriting the equation $ak^3+bk^2+ck$ with $k=1$, $k=2$, and $k=3$ (for which we know the corresponding results $1, 5, 14$) obtaining the following three equations:
$$a+b+c=1$$
$$8a+4b+2c=5$$
$$27a+9b+3c=14$$
The, we can solve the system, getting $a=1/3$, $b=1/2$, and $c=1/6$. This leads to
$$1/6\,(2k^3+3k+k)=1/6 \, k(k+1)(2k+1)$$
A: Approach 1 (fast): rewrite it as 
$$
a_k -a_{k-1} =k^2\\
\Delta a_k = k^2
$$
This is a telescoping sum on the left (check this!). If you sum from 1 to $n$ you get 
$$
a_n - a_1 = \sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6}
$$
The closed form expression can be obtained by using perturbation method or induction.
Approach 2 (hard and long). Use the generating function. It is $G(z) = \sum_{k=1}^{\infty} a_k z^k$. Then you need to solve 
$$
\frac{G(z)}{z} = G(z) + \sum_{k=1}^{\infty} k^2 z^k
$$
which was obtained by multiplying RHS and LHS by $z^k$ and summing over $k$. Now do the algebra and get an expression of the form 
$$
\sum_{k=1}^{\infty}a_k z^k = \sum_{k=1}^{\infty} \varphi(k) z^k
$$
You will find $\varphi(k)$ after some quick algebra. The closed-form solution will be 
$$
a_n = \varphi(n)
$$
A: $y_n=1+1/6\cdot n(n+1)(2n+1)-1=1/6\cdot n(n+1)(2n+1)$, because $\sum_{k=1}^n k^2=1/6\cdot n(n+1)(2n+1)$
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