Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
    
You asked almost the same question six hours ago. Apply the same method. –  Claude Leibovici Jul 6 at 13:38
    
If this is homework, please tag your post accordingly. –  Claude Leibovici Jul 6 at 13:42
    
It's just for practice. –  user3676752 Jul 6 at 13:45
    
See Faulhaber's formulas. –  Lucian Jul 6 at 14:09

3 Answers 3

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

$$1/6\,(2k^3+3k+k)=1/6 \, k(k+1)(2k+1)$$

share|improve this answer
    
Do you mind explaining how to get the three equations? –  user3676752 Jul 6 at 14:31
    
Hi, I edited my answer to better explain how to get the 3 equations. –  Anatoly Jul 6 at 14:43
    
I understand it now,thanks –  user3676752 Jul 6 at 14:44

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 $$ a_n = \varphi(n) $$

share|improve this answer

$y_n=1+1/6\cdot n(n+1)(2n+1)-1=1/6\cdot n(n+1)(2n+1)$, because $\sum_{k=1}^n k^2=1/6\cdot n(n+1)(2n+1)$

-

share|improve this answer
    
can u explain this part?I don't really understand how to get to this: ∑nk=1k2=1/6⋅n(n+1)(2n+1) –  user3676752 Jul 6 at 13:50
1  
Here yaikhom.com/2012/09/09/sum-of-squares.html is a derivation –  calculus Jul 6 at 14:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.