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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 :)

Daniel Andersson
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1 Answers1

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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

lab bhattacharjee
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