Question

Sa se calculeze:
a) 1+3+5+...+27
b) (2^10-1)(2^9-1)•...•(2^-9 -1)(2^-10 -1)

Answer 1

 

$\displaystyle\bf\\a) \\\\1+3+5+...+27=?\\\\\text{\bf Calculam numarul de termeni:}\\\\n=\frac{27-1}{2}+1=\frac{26}{2}+1=13+1=\boxed{\bf14}\\\\1+3+5+...+27=\frac{n(27+1)}{2}=\frac{14\cdot28)}{2}=14\cdot14=14^2=\boxed{\bf196}$

.

$\displaystyle\bf\\ b)\\\\(2^{10}-1)(2^9-1)...(2^{-9} -1)(2^{-10} -1)=\\\\=(2^{10}-1)(2^9-1)...(2^1-1)(2^0-1)(2^{-1}-1)...(2^{-9} -1)(2^{-10} -1)=\\\\\text{\bf Observam ca una din paranteze este zero : }~\boxed{\bf(2^0-1)=1-1=0}\\\\\implies(2^10-1)(2^9-1)...(2^-9 -1)(2^-10 -1)=\\\\=(2^{10}-1)(2^9-1)...(2^1-1)\underbrace{(2^0-1)}_{=0}(2^{-1}-1)...(2^{-9} -1)(2^{-10} -1)=\\\\=(2^{10}-1)(2^9-1)...(2^1-1)\times0\times(2^{-1}-1)...(2^{-9} -1)(2^{-10} -1)=\boxed{\bf0}\\\text{deoarece un factor este zero.}$

Answer 2

a)

Obs ca este o progresie aritmetica cu primul termen a1=1 si ratia r=a2-a1=3-1=2

an=a1+(n-1)r<=>27=1+2n-2<=>28=2n=>n=14

S14=(a1+a14)n/2=(1+27)*14/2=28*7=196

b)

(2^10-1)(2^9-1)•...•(2^1-1)(2^0-1)•...•(2^-9 -1)(2^-10 -1)

2^0-1=1-1=0 =>(2^10-1)(2^9-1)•...•(2^-9 -1)(2^-10 -1)=0