Home
Class 11
MATHS
" (5) "i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0...

" (5) "i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0

Promotional Banner

Similar Questions

Explore conceptually related problems

Show that: {i^(19)+((1)/(i))^(25)}^(2)=-4( ii) {i^(17)-((1)/(i))^(34)}^(2)=2i (iii) {i^(18)+((1)/(i))^(24)}^(3)=0 (iv) i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0 for all n in N.

Show that : (i) {i^(19) + (1/i)^(25)}^(2) = -4 " "(ii) {i^(17) - (1/i)^(34)}^(2) = 2i (iii) {i^(18) + (1/i)^(24)}^(3) = 0 " " (iv) i^n + i^(n+1) + i^(n +2) + i^(n + 3) = 0 , for all n in N .

If i= sqrt-1 and n is a positive integer , then i^(n) + i^(n + 1) + i^(n + 2) + i^(n + 3) is equal to

Prove that i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0 , for all n in N .

Prove that i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0 for all n inN

Sum of four consecutive powers of i(iota) is zero. i.e., i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0,forall n in I. If sum_(n=1)^(25)i^(n!)=a+ib, " where " i=sqrt(-1) , then a-b, is

Sum of four consecutive powers of i(iota) is zero. i.e., i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0,forall n in I. If sum_(n=1)^(25)i^(n!)=a+ib, " where " i=sqrt(-1) , then a-b, is

Sum of four consecutive powers of i(iota) is zero. i.e., i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0,forall n in I. If sum_(n=1)^(25)i^(n!)=a+ib, " where " i=sqrt(-1) , then a-b, is

Sum of four consecutive powers of i(iota) is zero. i.e., i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0,forall n in I. If sum_(n=1)^(25)i^(n!)=a+ib, " where " i=sqrt(-1) , then a-b, is