Prove, by using induction, $\frac{0^2+1^2+2^2+...+n^2}{n^2+n^2+n^2+...+n^2} = \frac{1}{3} + \frac{1}{6n}$

Question

Prove, by using induction, $\frac{0^2+1^2+2^2+...+n^2}{n^2+n^2+n^2+...+n^2} = \frac{1}{3} + \frac{1}{6n}$ for values of n>=1

The above summation can also be written as: $\frac{\sum_{i=0}^n i^2}{\sum_{i=0}^n n^2} = \frac{1}{3} + \frac{1}{6n}$

Here is how far I have gotten...

First show it is true for n=1,

$L.H.S =\frac{0^2+1^2}{1^2+1^2} = \frac{1}{2}, R.H.S. = \frac{1}{3} + \frac{1}{6*1} = \frac{1}{2}$

(Inductive Hypothesis) Assume it is true for n=k, then

$\frac{0^2+1^2+...+k^2}{k^2+k^2+...+k^2} = \frac{k(k+1)(2k+1)}{6(k^2)(k+1)}= \frac{1}{3} + \frac{1}{6k}$

using the property $\sum_{i=0}^n i^2 = \frac{k(k+1)(2k+1)}{6}$, as well as noticing the denominator in the above forumla is adding $k^2 (k+1)$ times

(Inductive Step)

Let n=k+1

$\frac{0^2+1^2+...+k^2+(k+1)^2}{(k+1)^2+...+(k+1)^2} = \frac{(k+1)(k+2)(2(k+1)+1)}{6(k+2)(k+1)^2}$ by the Inductive Hypothesis, then after cancelling like terms we get

$=\frac{2(k+1)+1}{6}=\frac{k+1}{3}+\frac{1}{6}$

Obviously, this is not the desired answer (I need the (k+1) in the other denominator somehow). I'm not sure where I'm going wrong. Any help is very much appreciated.

Answer 1

• For $n=1$, $\frac{\sum_{k=1}^1k^2}{(1+1)1^2}=\frac 1 2=\frac 1 3+\frac 1 6$
• Assuming that $\frac{\sum_{k=1}^nk^2}{(n+1)n^2}=\frac 1 3+\frac 1 {6n}$, we have that $$\sum_{k=1}^nk^2={(n+1)n^2}(\frac 1 3+\frac 1 {6n})$$ Then, \begin{align}\frac{\sum_{k=1}^{n+1}k^2}{(n+2)(n+1)^2}&=\frac{{(n+1)n^2}(\frac 1 3+\frac 1 {6n})+(n+1)^2}{(n+2)(n+1)^2}\\ &=\frac{{1 \over 6}(2n+3)(n+2)}{(n+2)(n+1)}\\ &=\frac 1 3+\frac 1 {6(n+1)}\end{align}