Show that `sum_(n=1)^(25) (i^(n) + i^(n +1)) = i-1`

The value of sum_(n=1)^(13) (i^n+i^(n+1)) , where i =sqrt(-1) equals (A) i (B) i-1 (C) -i (D) 0

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 Ndot

sum_(i=1)^(n) sum_(i=1)^(n) i is equal to

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

Evaluate sum_(n=1)^(13)(i^n+i^(n+1)), where n in Ndot

Evaluate sum_(n=1)^(13)(i^n+i^(n+1)), where n in Ndot

For any n positive numbers a_(1),a_(2),…,a_(n) such that sum_(i=1)^(n) a_(i)=alpha , the least value of sum_(i=1)^(n) a_(i)^(-1) , is

Prove that sum_(1lt=ilt) sum_(jlt=n)(i) = (n (n^2 - 1))/(6)

A_(1),A_(2), …. A_(n) are the vertices of a regular plane polygon with n sides and O ars its centre. Show that sum_(i=1)^(n-1) (vec(OA_(i))xxvec(OA)_(i+1))=(n-1) (vec(OA)_1 xx vec(OA)_(2))

A_1,A_2,..., A_n are the vertices of a regular plane polygon with n sides and O as its centre. Show that sum_(i=1)^n vec (OA)_i xx vec(OA)_(i+1)=(1-n)(vec (OA)_2 xx vec(OA)_1)