Question

a) i•i^2•...•i^10 =?
b) 1+i+i^2+...+i^10= ?
Îmi puteţi explica rezolvarea, vă rog?

Answer 1

$a)~i\cdot i^2\cdot...\cdot i^{10}=i^{1+2+...+10}=i^{55}$
$b)~a=1+i+i^2+...+i^{10} \\ \\ i\cdot a=i+i^2+i^3+...+i^{11} \\ \\ i\cdot a-a=i+i^2+i^3+...+i^{11}-1-i-i^2-...-i^{10} \\ \\ (i-1)a=i^{11}-1 \\ \\ a=\frac{i^{11}-1}{i-1}$

Answer 2

a)
i•i^2•...•i^10 =? => i•i^2•...•i^10 = -i

//folosim propietatile :
a^m•a^n=a^m+n ; i^2 =-1

i•i^2•...•i^10 =i^1+2+3...+10 ( 1+2+3+...10 = 55)
= i^55 =(i^54)•i = (i^2)^27*i = -i

a)
1+i+i^2+...+i^10=?
1+(i+i^2+...+i^10 )
i+i^2+...+i^10 este o progresie geometrica cu (ratia)q=i , b1= i , bn= i^10 , n = 10
S10 = b1(q^n - 1) / q-1 = i(i^10-1)/i-1 //( i^10 = -1 )
= i(-1-1)/i-1 = -2i/i-1(rationalizam) = -2i(i+1) / (i-1)(i+1) = -2i^2-2i /i^2-1 = 2-2i /-2 = 2(1-i) / -2 = i-1
deci , (i+i^2+...+i^10 ) = i-1 .Astfel suma devine:
1+i+i^2+...+i^10= 1+i+i-1 =2i