why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$ (Forgive me if it is a silly question)
When I was solving a puzzle, I observed a sequence
1, 1+8, 1+8+16, 1+8+16+24, 1+8+16+24+32.... 

is equals to
1,   9,     25,     49,    81.....

for which I see it as:
$(8 \times 0) +1, (8 \times 0 + 8 \times 1) +1, (8 \times 0 + 8 \times 1 + 8 \times 2) +1, (8 \times 1 + 8 \times 2 + 8 \times 3 + 8 \times 4) +1 ....$
equals to
$1^2, 3^2, 5^2, 7^2 ...$
and my brain stuck here and cannot find out the relationship between two series. 
Can someone give me a hint why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$

$$S_1=1=(2\cdot1-1)^2,\\
S_n-S_{n-1}=(2n-1)^2-(2n-3)^2=8n-8.$$

You can establish the formula using the well-known triangular numbers,
$$1+2+3+\cdots n=\frac{n(n+1)}2.$$
Then
$$S_n=8\frac{(n-1)n}2+1=4n^2-4n+1.$$
