Question

Mathematical Induction. [Number Theory]

1^2+3^2+5^2+...(2n-1)^2= n(2n-1)(2n+1)/3

Answer 1

I have the basis

Answer 2

I just need to work on the assume, and prove of the it.

Answer 3

Now i just need to prove it.

Answer 4

asume this is true for n=m then,

$$\sum\limits_{i=1}^m (2m-1)^2=\frac{m(2m+1)(2m-1)}{3}$$
add [2(m+1)-1]^2 to both sides,
$$\begin{align*}\sum\limits_{i=1}^m (2m-1)^2+[2(m+1)-1]^2&=\frac{m(2m+1)(2m-1)}{3}+[2(m+1)-1]^2\\
\sum\limits_{i=1}^{m+1} (2m-1)^2&=\frac{4m^3-m+12m^2+12m+3}{3}\\
&=\frac{(m+1)(4m^2+8m+3)}{3}\\
&=\frac{(m+1)(2m+3)(2m+1)}{3}\\
&=\frac{(m+1)[2(m+1)+1][2(m+1)-1)}{3}
\end{align*}$$
So, it is true for n=m+1

Answer 5

little typo in left side. it should be,
$$\sum\limits_{i=1}^{m}(2i-1)^2$$ not $$\sum\limits_{i=1}^{m}(2m-1)^2$$