# proof for $1=1^2$ ,$1+3=4=2^2$,$1+3+5=9=3^2$,$1+3+5+7=16=4^2$ [duplicate]

We notice that $1=1^2$ ,$1+3=4=2^2$,$1+3+5=9=3^2$,$1+3+5+7=16=4^2$ How we can generalize this and give a proof?Any ideas?

• How about using a simple induction? – Hamed Nov 22 '15 at 21:32

I have a fondness for this result as I discovered it myself in junior high.

My method was. Let $x = \sum_{i = 1}^n 2i - 1$. So $2x = \sum_{i = 1}^n 2i - 1 + \sum_{i=1}^n 2(n-i) + 1 = \sum_{i=1}^n [2i - 1 + 2n - 2i + 1] = \sum_{i=1}^n [2n ]=n * 2n = 2n^2$.

I was so proud of myself!

Then 40!!! years later I realize for the very first time that:

$(n + 1)^2 = n^2 + 2n + 1$ which means $(n+1)^2 - n^2 = 2n + 1$... this is the n-th odd number! So every n+1 the perfect square is the nth odd number larger than the nth perfet square, and by induction every perfect square is the sum of the consecutive odd numbers.

I had never noticed that before!

It was a dope slap 40 years in the making!

HINT: the difference between consecutive squares is $$(x+1)^2-x^2=(x^2+2x+1)-x^2=2x+1...$$

$$S_n=1+3+5+...+(2n-1)+(2n-3)+(2n-5)\\S_n=(2n-1)+(2n-3)+(2n-5)+...+3+2+1$$ sum of them is $2s_n$ $$2s_n=(1+2n-1)+(3+2n-3)+(5+2n-5)+...+(2n-3+3)+(2n-1+1)=\\(2n)+(2n)+...+(2n)$$ $n$ terms,so $$2s_n=(1+2n-1)+(3+2n-3)+(5+2n-5)+...+(2n-3+3)+(2n-1+1)=\\(2n)+(2n)+...+(2n)=2n \times n\\2s_n=2n^2\\s_n=n^2$$ another approach is visual proof : HINT: Consider how many elements are in each sum and what their average value will be. For example, 1+3 has 2 elements with an average of 2.

See if you can figure it out with that.