Use the principle of induction to show $2+6+18+\ldots+2\cdot 3^{n-1}=3^n-1$

Question

Show that $$2+6+18+\ldots+2\cdot 3^{n-1}=3^n-1$$

Proving the base case when $n=1$:
$2\cdot3^{1-1}=3^1-1\Leftrightarrow 2=2$

Now doing the induction:
$2\cdot 3^{(n+1)-1}=3^{n+1}-1$
$2\cdot 3^n=3^{n+1}-1$

But I'm kinda stuck here and don't know what to do next

Answer 1

You should start by using proper formulae:

Show that $$\sum_{i=1}^n 2\cdot 3^{i-1} = 3^n - 1$$

Base case: $2\cdot 3^0 = 2 \checkmark$
Induction hypothesis: $$\sum_{i=1}^n 2\cdot 3^{i-1} = 3^n - 1$$ Induction step: Show that $$\sum_{i=1}^{n+1} 2\cdot 3^{i-1} = 3^{n+1} - 1$$ Now the LHS is equal to $$\sum_{i=1}^n 2\cdot 3^{i-1} + 2 \cdot 3^{(n+1)-1} \stackrel{\text{i.h.}}= 3^n-1 + 2\cdot 3^n = 3\cdot 3^n - 1 = 3^{n+1} - 1$$ as claimed. $\Box$

More generally:

$$\sum_{k=0}^{n-1} (p-1) p^k = p^n - 1 \qquad \forall\ p, n$$

Answer 2

you must calculate $$3^{n}-1+2\cdot 3^n=3^{n+1}-1$$ this is equivalent to $$2\cdot 3^n+3^n=3^{n+1}$$ $$2\cdot 3^n=3^n(3-1)$$ which is true.

Answer 3

The base case is done, now you assume the Induction Hypothesis and prove it for $"n+1"$.

The way you are stating the problem is wrong, why? Because you want to prove that:

$\sum_{i=0}^{n-1} 3^{i-1}\cdot 2 = 3^{n}-1$

Now for the Inductive Step, use your induction hypothesis. Using it will give you:

$\sum_{i=0}^{n} 3^{i-1}\cdot 2 = 3^{n}-1 + 3^{n} \cdot 2 = 3\cdot 3^{n}-1 = 3^{n+1}-1$

And that's it.

Answer 4

You need to prove by induction that $\sum\limits_{k=0}^{n-1}2\cdot3^k=3^n-1$

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First, show that this is true for $n=1$:

• $\sum\limits_{k=0}^{0}2\cdot3^k=3^1-1$

Second, assume that this is true for $n$:

• $\sum\limits_{k=0}^{n-1}2\cdot3^k=3^n-1$

Third, prove that this is true for $n+1$:

• $\sum\limits_{k=0}^{n}2\cdot3^k=\sum\limits_{k=0}^{n-1}2\cdot3^k+2\cdot3^n$

• $\sum\limits_{k=0}^{n-1}2\cdot3^k+2\cdot3^n=3^n-1+2\cdot3^n$ assumption used here

• $3^n-1+2\cdot3^n=3\cdot3^n-1$

• $3\cdot3^n-1=3^{n+1}-1$