By induction, show that for ∀n∈N, it is true that:

Question

$$\sum_{i=1}^n 2^i=2+2^2+2^3+.....+ 2^n=2(2^{n}-1)$$ Any help/explanations would be REALLY appreciated.

Also in the same vein:

By induction, show that $$∀n∈\mathbb N: 11^{n+2} + 12^{2n+1}$$ is divisible by $133$.

Thank you so much.

Answer 1

Going for the second one.

Start at $n = 1$: $$ 11^3 + 12^3 = 3059 = 23 \cdot 133 $$

Induction step from $n$ to $n+1$: \begin{align} 11^{n+3} + 12^{2n+3} &= 11 \cdot 11^{n+2} + 12^2 \cdot 12^{2n+1} \\ &= 11(11^{n+2} + 12^{2n+1})+133 \cdot 12^{2n+1} \end{align} where the first term is a multiple of $133$ by the validity of the statement for $n$ and the second term is obviously a multiple of $133$, so the statement for $n+1$ is a multiple of $133$ as well.

Answer 2

$$p(1): i=1 \rightarrow 2^i=2(2^1-1)\\p(k): 2^1+2^2+2^3+...+2^k=2(2^{k}-1)\\p(k+1): 2^1+2^2+2^3+...+2^k+2^{k+1}=2(2^{k+1}-1)\\ $$now add the last term of p(k+1) to p(k) $$p(k)+2^{k+1}\\2^1+2^2+2^3+...+2^k+2^{k+1}=2(2^{k}-1)+2^{k+1}\\2(2^{k}-1)+2^{k+1}=2(2^{k}-1)+2*2^{k}=2(2^k-1+2^k)\\=2(2*2^{k}-1)=2(2^{k+1}-1) $$

Answer 3

Since mvw have answerd second question, I'll solve only first one

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Using Induction:

Let $$s_n=\sum_{i=1}^n 2^i=2+2^2+2^3+.....+ 2^n=2(2^{n}-1)$$

For $n=k$

$$s_k=\sum_{i=1}^k 2^i=2+2^2+2^3+.....+ 2^k=2(2^{k}-1)$$

For $n=k+1$

$$s_{k+1}=\sum_{i=1}^{k+1} 2^i=2+2^2+2^3+.....+ 2^{k+1}=2(2^{k}-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2(2^{k+1}-1)$$

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Without Induction

$$\begin{align} s_n&=2+2^2+2^3+.....+ 2^n\\ 2s_n&=\,\,\,\,\,\,\,\,\,\,2^2+2^3+.....+ 2^n+2^{n+1}\tag{1}\\ 2s_n-s_n&=2^{n+1}-2=2(2^n-1)\tag{2}\\ s_n&=2(2^n-1)\\\end{align}$$