Question

11^(1+3+5+...+31)=11^n^2
2^(5+12+19+...+75)=2^11n
URGENT DAU COROANA

Answer 1

   
[tex]\displaystyle\\ 1)\\ 11^{(1+3+5+...+31)}=11^{n^2}\\\\ 1+3+5+...+31 = n^2\\\\ \texttt{Aflam numarul de termeni din sir.}\\\\ k = \frac{31-1}{2}+1 = \frac{30}{2}+1 =15+1 = 16 \\\\ 1+3+5+...+31 = \frac{16(31+1)}{2} =\frac{16 \times 32}{2} = 16\times 16 = 16^2\\\\ 16^2 = n^2 \boxed{\bf n = 16} [/tex]


[tex]\displaystyle\\ 2)\\ 2^{(5+12+19+...+75)}=2^{11n}\\\\ (5+12+19+...+75) = 11n\\\\ \texttt{Calculam numarul de termeni ai sirului.}\\\\ k = \frac{75-5}{7}+1 = \frac{70}{7}+1 = 10+1=11\\\\ 5+12+19+...+75= \frac{11(75+5)}{2} =\frac{11 \times 80}{2} = 11 \times 80\\\\ 11 \times 80 = 11 \times n~~~\Big|~:11 \\\\ \boxed{\bf n = 80}[/tex]

Answer 2

11^(1+3+5+...+31)=11^n^2
1+3+5+...+31= (1+31)•15/2= 32•16/2= 16•16
=>11^(1+3+5+...+31)=11^(16•16)=11^16^2
=> n=16

2^(5+12+19+...+75)=2^11n
5+12+19+...+75= (5+75)•11/2= 80•11/2= 40•11
=> 2^(5+12+19+...+75)=2^11•40
=> n=40