## Loser66 one year ago 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

1. anonymous

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

2. Loser66

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?

3. Loser66

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

4. Loser66

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

5. Loser66

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?

6. Loser66

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

7. ganeshie8

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

8. Loser66

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

9. Empty

$\sum_{i=1}^{i=n} a^i$ 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

10. ganeshie8

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.

11. ganeshie8

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

12. Loser66

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. :(

13. ganeshie8

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 : $$\sum_{j=1-1}^{5-1} 3(j+1)^2=3\sum_{j=0}^4 (j+1)^2 =3(1^2+2^2+3^2+4^2+5^2)$$

14. ganeshie8

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

15. Loser66

Got this part :) Now other example

16. Loser66

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

17. ganeshie8

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

18. Loser66

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