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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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

Mathematics
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\[\sum_{i=0}^{n}(i)^(n-1)\]
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?
@AliLnn cannot be n as it is on original one. :(

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Other answers:

sorry, *\(a^{i-1}\)
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?
At that moment , the upper limit is n +1, not n-1, That makes me confused
\(\sum_{i=1}^n a^i\) Let \(j=i-1\) : \(i=1\implies j=0\) \(i=n\implies j=?\)
@ganeshie8 j = n-1, but how to argue for my example above?
\[\sum_{i=1}^{i=n} a^i\] 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
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.
maybe try contrasting below two identical sums : \(\sum\limits_{i=1}^6e^i\) and \(\sum\limits_{i=0}^5e^{i+1}\)
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. :(
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 : \(\sum_{j=1-1}^{5-1} 3(j+1)^2=3\sum_{j=0}^4 (j+1)^2 =3(1^2+2^2+3^2+4^2+5^2)\)
Notice that both the series are one and same and they match term by term
Got this part :) Now other example
\(\sum_{i =1}^5 3^i\)
\(\sum_{i =1}^5 3^i\) substitute \(j=i-1\), the sum can be rewritten as \(\sum_{j =1-1}^{5-1} 3^{j+1}\)
Thank you so so much. I got it now. :)

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