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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?

wolfram-alpha result

Hamada Al
  • 938
Carlos
  • 87
  • First... do you mean to have the summands be $n^l$ or $2^l$? Where did the $2$ come from? Second, what do the last few terms in $\sum\limits_{l=k}^n 2^l$ look like? Double check for yourself... are there $n$ terms in the summation? or is it just that the last index should be $n$ and there could be fewer than $n$ terms in the summation? – JMoravitz May 10 '22 at 21:33
  • @JMoravitz, Check again it was a typo – Carlos May 10 '22 at 21:35
  • $a_n=a_1r^{n-1}$ is the last term of the summation – Carlos May 10 '22 at 21:40

2 Answers2

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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.

Hamada Al
  • 938
  • This seems correct thanks, however i was wondering how can someone reach the conclusion of changing the index without taking a look to the answer? I forgot to mention that the exercise tells you that $1 \leq k \leq n$, could that possibly lead you to the conclusion above? Sorry i'm new with summations and series. – Carlos May 11 '22 at 02:14
  • To begin the index from zero, you should subtract $k$ from $l$ and you get new index $m=l-k$ from $0$ to $n-k$, but $l=m+k$, thus it becomes $\sum_{m=0}^{n-k}2^{m+k}$. – Hamada Al May 11 '22 at 09:15
  • Now you can replace the dummy variable $\sum_{m=0}^{n-k}2^{m+k}=\sum_{l=0}^{n-k}2^{l+k}$ – Hamada Al May 11 '22 at 09:17
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$$\sum_{l = k}^n \; \text{gives} \; n-k + 1 \; \text{terms}$$ $$2^k + 2^{k+1} + 2^{k+2} + ... + 2^{k}\cdot2^{n-1} \; \text{gives} \; n \; \text{terms} \tag{1}$$

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