$\sum_{i=1}^n a_i=\sum_{i=0}^{n-1} a_{i+1} $

Question

how to prove that:

$$\sum_{i=1}^n a_i=\sum_{i=0}^{n-1} a_{i+1} $$ where $a_i$ is any statement that dependent on $i$.

I was trying to prove the binomial theorem by induction,So I needed this property. However,i remember that in Spivak's calculus answer book that if there was any thing like that,then the solution would be like this:

$$\sum_{i=1}^n a_i=\sum_{k=0}^{k-1} a_{k} \quad\text{(let }k=i-1)$$

That's why I ask this relatively easy question.

it should be obvious. if you are asking to prove it rigorously, you should use induction. — mathworker21, Sep 18 '16 at 23:15

score 2 · Answer 1 · answered Sep 18 '16 at 23:09

2

Consider the sum $$\sum_{i = 1}^n a_i$$ If we use the index substitution $i = k + 1$, we get for the lower index $$i = k + 1 = 1\Leftrightarrow k = 0$$ and for the upper index $$i = k + 1 = n \Leftrightarrow k = n - 1$$ Thus $$\sum_{i = 1}^n a_i = \sum_{k = 0}^{n-1} a_{k+1}$$ Just substitute $k = i$ and you are done.

answered Sep 18 '16 at 23:09

TheGeekGreek

7,869

yes. the point where you substitute $k$=$i$ although $i$=$k$+1. I think it is true because the sigma takes the numbers and input it no matter the variable is. I'm sorry my speech is full of such ambiguities currently ,but I really want to understand this point as it is my main reason why I asked this question. – someone Sep 19 '16 at 01:13

score 1 · Answer 2 · answered Sep 18 '16 at 23:17

You can write

$$\sum_{i=1}^{n}a_i=\sum_{1\le i\le n}a_i$$

Then

$$1\le i\le n\iff 1-1\le i-1\le n-1$$

Now if we change $i-1=k$ then

$$1\le i\le n\iff 0\le k\le n-1$$

Then

$$\sum_{i=1}^{n}a_i=\sum_{1\le i\le n}a_i=\sum_{0\le k\le n-1}a_{k+1}=\sum_{k=0}^{n-1}a_{k+1}$$

score 0 · Answer 3 · answered Sep 18 '16 at 23:12

0

It's simply true. Look at

$$ \sum_{i=1}^n a_n = a_1+a_2+\cdots +a_n=a_{0+1}+a_{1+1}+a_{(n-1)+1} = \sum_{i=0}^{n-1}a_{i+1} $$

answered Sep 18 '16 at 23:12

NewbcakeDK

71

$\sum_{i=1}^n a_i=\sum_{i=0}^{n-1} a_{i+1} $

3 Answers3