Induction: Sum of the squares of 6 consecutive natural numbers

Question

Define for every natural n:

$$ a_{n}=\sum\limits_{i=0}^{5}(n+i)^2$$

in other words, $\ a_n$ is the sum of the squares of 6 consecutive natural numbers, the first number is $n^2$ and the last is $(n+5)^2$.

Prove (by induction) that for every natural $n$ that $a_n$ has remainder $7$ mod $12$.

What I did so far:

$$(n^2+(n+1)^2+(n+2)^2+(n+3)^2+(n+4)^2+(n+5)^2)\mod 12=\\((n+1)^2+(n+2)^2+(n+3)^2+(n+4)^2+(n+5)^2+(n+6)^2)\mod 12$$

I have no idea how to continue, or if it's a good start at all...

$a_n$ gives a remainder of 7 when divided by 12. That means that $a_n \mod 12 = 7$ — Shahar, Apr 15 '14 at 22:22
thanks!........I want to add that the OP(me) has no idea about induction, any help appreciated — Seth Keno, Apr 15 '14 at 22:25
You could see Arturo Magidin's answer to this question for an introduction to induction. — Ross Millikan, Apr 15 '14 at 22:30

score 5 · Answer 1 · answered Apr 15 '14 at 22:23

5

Hint: For induction, you need to prove a base case. Here it would be to verify that $a_0\equiv 7 \pmod {12}$, which you can do just by computation. Then assume $a_k\equiv 7 \pmod {12}$ and write $a_{k+1}=\sum\limits_{i=0}^{5}(k+1+i)^2=a_k-k^2+(k+6)^2$ Now if you can show $(k+6)^2-k^2 \equiv 0 \pmod {12}$ you are home.

answered Apr 15 '14 at 22:23

Ross Millikan

374,822

thanks for the answer, I'll try that – Seth Keno Apr 15 '14 at 22:26
any hint how to show that $\ (k+6)^2-k^2 \equiv 0 \pmod{12}$? – Seth Keno Apr 15 '14 at 22:50
If you expand it out, the $k^2$ terms cancel and each of the other terms has a factor $12$ in it. – Ross Millikan Apr 15 '14 at 22:54

Induction: Sum of the squares of 6 consecutive natural numbers

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