Summer Clouds moderators
24/07/2017 at 09:02-
Sorry, can I answer again?
\(2^1.2^2.2^3...2^n=2^{5050}\) => \(2^{1+2+3+...+n}=2^{5050}\)
=> \(1+2+3+...+n=5050\)
In the series, there are: \(\left(n-1\right):1+1\) numbers.
=> \(n.n+1:2=5050\)
=> \(n.n+1=5050.2=10100\)
We saw that: \(100.101=5050\) => \(n=100\)
Selected by MathYouLike -
\(2^1.2^2.2^3...2^n=2^{5050}\) => \(2^{1+2+3+...+n}=2^{5050}\)
=> \(1+2+3+...+n=5050\)
=> \(n=100\)