5+13+21+...+(8n-3)=n(4n+1)
n=1
L.H.S.=(8(1-3))=5
R.H.S.=(1)(4(1)+1)=5
Thus, L.H.S.=R.H.S. for n\geq{1}
Prove for (n+1), 5+11+19+...+(8n-3)+[8(n+1)-3=(n+1)[4(n+1)+1)]
\underbrace{5+11+19+...+(8n-3)}_\text{n(4n+1)}+[8(n+1)-3]=(n+1)[4(n+1)+1)]
n(4n+1)+[8(n+1)-3]=4n^2+9k+5
4n^2+9n+5=4n^2+9k+5
Hence...