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$$3+3 \cdot 5+3 \cdot 5^2+ \cdots +3 \cdot 5^n =\frac{3 \cdot (5^{n+1} -1)}{4}$$

basic step: n=1 $\quad 3+3 \cdot 5 = 3 \cdot (5^{1+1} -1)/4 \iff 18=18 \;$,  true

assume: $\quad3+3\cdot 5+3\cdot 5^2+...3\cdot 5^k =3\cdot (5^{k+1} -1)/4$

then:

$3\cdot (5^{n+1} -1)/4 +3 \cdot 5^{n+1} = 3\cdot (5^{n+2} -1)/4$

$3\cdot (5^{n+1} +4\cdot 5^{n+1} -1)/4 = 3\cdot (5^{n+2} -1)/4.$

Could someone explain how the left side can be done to the same as right side with all the steps? It's part of an inductive proof but don't know how to finish it.

dxiv
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Deloss
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  • I think you're missing something. That equation is not true. Maybe try checking your inductive hypothesis. – Zaros Nov 18 '16 at 16:17
  • It should be. It's from a math book answers and anyway if you replace n with some random number it's equal. that's not real proof but it should be correct anyway. – Deloss Nov 18 '16 at 16:21

1 Answers1

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You were one small step away. All that's left to do in the last expression is note that $$5^{n+1} + 4 \cdot 5^{n+1} = (1 + 4)\cdot 5^{n+1} = 5 \cdot 5^{n+1} = 5^{n+2}$$

which then gives the sought equality:

$$\frac{3\cdot (5^{n+2} -1)}{4} = \frac{3\cdot (5^{n+2} -1)}{4}$$

A few more notes...

  • You could have started the induction with $n=0$ as the base case: $3 \cdot 5^0 = 3\cdot(5^1-1)/4$.

  • You use $k$ in the "assume" part, but $n$ in the induction step.

  • The left hand side is just 3 times the sum of a geometric progression, so without induction:

$$3\cdot(1 + 5 + \cdots + 5^n) = 3\cdot \frac{5^{n+1}-1}{5-1}$$

dxiv
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