`1+i^(10)+i^(110)+i^(1000)`

Prove that: (i) 1+i^(10)+i^(100)-i^(1000)=0 (ii) i^(107)+i^(112)+i^(117)+i^(122)=0 (iii) (1+i^(14)+i^(18)+i^(22)) is real number.

Prove that: (i) 1+i^(2)+i^(4)+i^(6)=0 (ii) 1+i^(10)+i^(100)+i^(1000)=2 (iii) i^(104)+i^(109)+i^(114)+i^(119)=0 (iv) 6i^(54)+5i^(37)-2i^(11)+6i^(68)=7i (v) (i^(592)+i^(590)+i^(588)+i^(586)+i^(584))/(i^(582)+i^(580)+i^(578)+i^(576)+i^(574))=-1

Write the following in the form x+iy: (i) i+i^(2)+i^(3)+i^(4) (ii) i^(4)+i^(8)+i^(12)+i^(16) (iii) i+i^(5)+i^(9)+i^(13) (iv) i^(9)+i^(10)+i^(11)+i^(12) .

Write the value of i+i^(10)+i^(20)+i^(30)

Show that 1+i^(10)+i^(20)+i^(30) is a real number.

Simplify the following : (i) 1+ i^(5)+i^(10)+i^(15) (ii) (1+i)^(4)+(1+(1)/(i))^(4) (iii) i^(n)+i^(n+1)+i^(n+2)+i^(n+3)

Find the values of following expressions: i^(49)+i^(68)+i^(89)+i^(110) (ii) i^(30)+i^(80)+i^(120) (iii) i^+i^2+i^3+i^4 (iv) i^5+i^(10)+i^(15) (v) (i^(592)+i^(590)+i^(586)+i^(584))/(i^(582)+i^(580)+i^(576)+i^(574)) (vi) 1+i^2+i^4+i^6+i^8+doti^(20) (vii) (1+i)^6+(1-i)^3

If i^(2)=1, then the sum i+i^(2)+i^(3)+ upto 1000 terms is equal to 1 b.-1 c.i d.0

The compound interest on a certain sum is given by C.I.=P(1+(R)/(100))^(n)-P . Find C.I. when P=rs1000,R=10% P.a., and n=2. The following steps are involved in solving the above problem. Arrange them in sequential order. (A) therefore C.I.=rs210 (B) 1000((11)/(10))((11)/(10))-1000=1210-100 (C) Given Cl=P(1+(R)/(100))^(n)-P,P=rs1000 , R=10% p.a., and n=2 (D) C.I. = 1000+(1+(10)/(100))^(2)-1000