Make it clear, please.
I confused. How to turn it to i = 0
$\sum_{i=1}^n a^i$ to $\sum_{i=0}^{??}$
Please, help

Question

anonymous · Answer

$$\sum_{i=0}^{n}(i)^(n-1)$$

loser66 · Answer

I was taught that when we replace i =1 from i =0, whenever I saw i, subtract 1
Hence the lower limit is i=0, the summand is $a^{n-1} $ . How about the upper limit?

loser66 · Answer

@AliLnn  cannot be n as it is on original one. :(

loser66 · Answer

sorry, *$a^{i-1}$

loser66 · Answer

Ex: $\sum_{i=1}^5 i =1 +2+3+4+5$ when I turn it to i =0, the sum must be
$\sum_{i=0}^6 i = 0 +1+2+3+4+5$ to make it the same with the original one, right?

loser66 · Answer

At that moment , the upper limit is n +1, not n-1, That makes me confused

ganeshie8 · Answer

$\sum_{i=1}^n a^i$

Let $j=i-1$ :
$i=1\implies j=0$
$i=n\implies j=?$

loser66 · Answer

@ganeshie8  j = n-1, but how to argue for my example above?

empty · Answer

$$\sum_{i=1}^{i=n} a^i$$

You want the lower bound to be 0, so we do algebra on the bottom part since it is but a simple equation!

 $$\sum_{i-1=0}^{i=n} a^i$$

Now we see our substitution clearly:

$$i-1=j$$

Rearrange this to plug in our new value everywhere:
$$i=j+1$$
Goes in:

$$\sum_{i-1=0}^{i=n} a^i=\sum_{j=0}^{j+1=n} a^{j+1}=\sum_{j=0}^{j=n-1} a^{j+1}$$

This is what you must do and nothing more, it is just tiny equations on top of and below a funny greek letter. :P

ganeshie8 · Answer

your example just happens to be a special case : 


0 + anything  = anything

that doesn't mean, we don't have two terms on the left hand side.

ganeshie8 · Answer

maybe try contrasting below two identical sums :
 $\sum\limits_{i=1}^6e^i$ and  $\sum\limits_{i=0}^5e^{i+1}$

loser66 · Answer

Ex2: $\sum_{i=1}^5 3i^2=3\sum_{i=1}^5 i^2 =3(1^2+2^2+3^2+4^2+5^2)$ the first element is 3, the last one is 75
If I want to turn it from 0 to ??
the sum must stay as it is. right?$\sum_{i =0 }^4 3(i^2) = 3(0^2+ 1^2+2^2+3^2 +4^2)$
Surely, the sum doesn't stay as it is. :(

ganeshie8 · Answer

I see, you must do the substitution inside also. : 

$\sum_{i=1}^5 3i^2=3\sum_{i=1}^5 i^2 =3(1^2+2^2+3^2+4^2+5^2)$

substitute $j=i-1$, the sum is same as :
$\sum_{j=1-1}^{5-1} 3(j+1)^2=3\sum_{j=0}^4 (j+1)^2 =3(1^2+2^2+3^2+4^2+5^2)$

ganeshie8 · Answer

Notice that both the series are one and same and they match term by term

loser66 · Answer

Got this part :)
Now other example

loser66 · Answer

$\sum_{i =1}^5 3^i$

ganeshie8 · Answer

$\sum_{i =1}^5 3^i$ 

substitute $j=i-1$, the sum can be rewritten as
$\sum_{j =1-1}^{5-1} 3^{j+1}$

loser66 · Answer

Thank you so so much. I got it now. :)