Question

2 la puterea 1+2+3+...+16= 4 la n+ 7 va rog! cum se face??

Answer 1

 [tex]Probabil\;ca\;avem:\;\\ 2^{1+2+3+...+16}=4^{n+7}\\ (1)...\;1+2+3+...+16=\frac{16*17}{2}=8*17=136\surd\surd\\ (2)...\;4^{n+7}=(2^2)^{n+7}=2^{2n+14}\;;\\ (3)...\;din \;egalitatea \;initiala\;\rightarrow\;2n+14=136\;adica\;n=61[/tex]