<s>[INST] 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 $(2n-1)^2$
 [/INST] $(n+1)^2 = n^2+(2n+1) $ so by induction $n^2= \sum 2k-1$
And for just the odd squares. $(2n+1)^2= 1 + \sum_{k=2;+2}^{2n} [(2k - 1)+(2k+1) ]= 1 +4\sum_{j=1}^{n }2j=1+8\sum j$
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