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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
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{i=0}^{n}(i)^(n1)\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I 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 \(a^{n1} \) . How about the upper limit?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@AliLnn cannot be n as it is on original one. :(

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Ex: \(\sum_{i=1}^5 i =1 +2+3+4+5\) 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
 one year ago
Best ResponseYou've already chosen the best response.1At that moment , the upper limit is n +1, not n1, That makes me confused

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5\(\sum_{i=1}^n a^i\) Let \(j=i1\) : \(i=1\implies j=0\) \(i=n\implies j=?\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 j = n1, but how to argue for my example above?

Empty
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{i=1}^{i=n} a^i\] You want the lower bound to be 0, so we do algebra on the bottom part since it is but a simple equation! \[\sum_{i1=0}^{i=n} a^i\] Now we see our substitution clearly: \[i1=j\] Rearrange this to plug in our new value everywhere: \[i=j+1\] Goes in: \[\sum_{i1=0}^{i=n} a^i=\sum_{j=0}^{j+1=n} a^{j+1}=\sum_{j=0}^{j=n1} a^{j+1}\] 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
 one year ago
Best ResponseYou've already chosen the best response.5your 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
 one year ago
Best ResponseYou've already chosen the best response.5maybe try contrasting below two identical sums : \(\sum\limits_{i=1}^6e^i\) and \(\sum\limits_{i=0}^5e^{i+1}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Ex2: \(\sum_{i=1}^5 3i^2=3\sum_{i=1}^5 i^2 =3(1^2+2^2+3^2+4^2+5^2)\) 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
 one year ago
Best ResponseYou've already chosen the best response.5I 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 \(j=i1\), the sum is same as : \(\sum_{j=11}^{51} 3(j+1)^2=3\sum_{j=0}^4 (j+1)^2 =3(1^2+2^2+3^2+4^2+5^2)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Notice that both the series are one and same and they match term by term

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Got this part :) Now other example

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5\(\sum_{i =1}^5 3^i\) substitute \(j=i1\), the sum can be rewritten as \(\sum_{j =11}^{51} 3^{j+1}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Thank you so so much. I got it now. :)
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