Finding the sum of finite geometric series

Question

I'm doing the following summation $\sum_{l=k}^{n}2^l$

$\sum_{l=k}^{n}2^l = 2^k + 2^{k+1} + 2^{k+2} + \ldots+ 2^{n-1} + 2^{n}$

$S_n=a_1\dfrac{1-r^n}{1-r} \therefore S_n=2^k\dfrac{1-(2)^n}{1-2} = 2^{k+n}-2^k$

But my final result seems to be incorrect compared to the one obtained within the calculator

Am i doing something wrong when using the formula above?

Answer 1

The following summation is

$$\sum_{l=k}^{n}2^l = 2^k + 2^{k+1} + 2^{k+2} + \ldots+ 2^{n-1} + 2^{n}$$

can be written as

$$\sum_{l=0}^{n-k}2^{l+k} = 2^k + 2^{k+1} + 2^{k+2} + \ldots+ 2^{n-1} + 2^{n}$$

by shifting the index $k$-times, which is a series with $n−k+1$ terms

$S_{n-k+1}=a_1\dfrac{1-r^{n-k+1}}{1-r} \therefore S_{n-k+1}=2^k\dfrac{1-(2)^{n-k+1}}{1-2} = 2^{n+1}-2^k$

This result seems to be correct compared to the one obtained within the calculator.

Answer 2

$$\sum_{l = k}^n \; \text{gives} \; n-k + 1 \; \text{terms}$$ $$2^k + 2^{k+1} + 2^{k+2} + ... + 2^{k}\cdot2^{n-1} \; \text{gives} \; n \; \text{terms} \tag{1}$$

To write $(1)$ using sigma notation, simply write $$\sum_{l = k}^{n + k - 1}2^{l} \; \text{or} \sum_{l=0}^{n-1} 2^{k + l}$$