<s>[INST] Show that $\sum_{k=1}^{n}(5^{2k}-5^{2k-1})=\frac{5^{2n+1}-5}{6}$, for n = 1,2,3, ... Show that
$\sum_{k=1}^{n}(5^{2k}-5^{2k-1})=\frac{5^{2n+1}-5}{6}$, for n = 1,2,3, ...
n = 1
$LHS = 5^2 - 5 = 20$
$RHS =\frac{ 5^{3} - 5}{6} = \frac{120}{6} = 20$
n = p 
$LHS_{p} = (5^{2(1)}-5^{2(1)-1}) + (5^{2(2)}-5^{2(2)-1})+....+ (5^{2(p)}-5^{2(p)-1}) $
$RHS_{p}  =\frac{ 5^{2(p)+1} - 5}{6} $
n = p + 1 
$LHS_{p+1} = (5^{2(1)}-5^{2(1)-1}) + (5^{2(2)}-5^{2(2)-1})+....+ (5^{2(p)}-5^{2(p)-1})+ (5^{2(p+1)}-5^{2(p+1)-1}$
$RHS_{p+1}  =\frac{ 5^{2(p+1)+1} - 5}{6} $
Show that: $ RHS_{p+1}  = RHS_{p} + 5^{2(p+1)}-5^{2(p+1)-1} $
$\frac{ 5^{2(p+1)+1} - 5}{6}  = \frac{ 5^{2(p)+1} - 5}{6}  + 5^{2(p+1)}-5^{2(p+1)-1} $ 
And here's where I am stuck :)
 [/INST] Hint:
$$5^{2k}-5^{2k-1}=5^{2k-1}(5-1)=4\cdot5^{2k-1}$$
So, we have
$$\sum_{k=1}^n4\cdot5^{2k-1}$$ which is a geometric Series
Use Induction proof dealing with geometric series
</s>