# Working on this summation proof

$$\sum_{i=1}^n t^{i-1}$$ I am stuck with the proof of this equality.

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Hint: Write out the summation $$\sum_{i=1}^n t^{i-1}=1+t+t^2+\cdots+t^{n-1}$$ Now multiply by $1-t$, and observe which terms will cancel and which will survive.

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Use induction on $n\in \mathbb N, n\ge1$ with $P(n):=\sum_{i=1}^n t^{i-1}=\frac{1-t^n}{1-t}$: For $n=1$ you have: $$P(1): 1=\frac{1-t}{1-t}$$ which is correct. Suppose that $P(n)$ is true, now check it for $n+1$. You have $$P(n+1): \sum_{i=1}^{n+1} t^{i-1}=\sum_{i=1}^n t^{i-1}+t^{n+1}=\frac{1-t^n}{1-t}+t^{n+1}=\frac{1-t^{n+1}}{1-t}$$ So the statement is true for any $n\in\mathbb N$.
Use perturbation method from Concrete Mathematics: $$S_n=\sum_{k=1}^{n}t^{i-1}\\ S_{n+1}=S_n + t^{n}=\sum_{k=1}^{n}t^{k} + 1=t\sum_{k=1}^{n}t^{k-1}+1=t S_n + 1$$ After the algebra you get $$S_n=\frac{1-t^{n}}{1-t}$$