(i)(1^(3)+2^(3)+3^(3))^((1)/(2))

Convert the following in the form of (a+ib) : (i) (1+i)^(4) (ii) (-3+(1)/(2)i)^(3) (iii) (1-i)(3+4i) (iv) (1+i)(1+ 2i)(1+ 3i) (v) (3+5i)/(6-i) (vi) ((2+3i)^(2))/(2+i) (vii) ((1+ i)(2+i))/((3+i)) (viii) (2-i)^(-3)

Convert the following in the form of (a+ib) : (i) (1+i)^(4) (ii) (-3+(1)/(2)i)^(3) (iii) (1-i)(3+4i) (iv) (1+i)(1+ 2i)(1+ 3i) (v) (3+5i)/(6-i) (vi) ((2+3i)^(2))/(2+i) (vii) ((1+ i)(2+i))/((3+i)) (viii) (2-i)^(-3)

I:2[((1)/(3))+(1)/(3)((1)/(3))^(3)+(1)/(5)((1)/(3))^(5)+...]=log_(e)2 II:2[((1)/(2))+(1)/(3)((1)/(2))^(3)+(1)/(5)((1)/(2))^(5)+...]=log_(e)2

(-2-i(1)/(3))^(3)

Find the following sums : (i) 1 + 2 + 3 + …+ 30 (ii) 1^(2) + 2^(2) + 3^(2) + …+35^(2) (ii) 1^(3) + 2^(3) + 3^(3) +…+ 20^(3) .

Find the following sums : (i) 1 + 2 + 3 + …+ 30 (ii) 1^(2) + 2^(2) + 3^(2) + …+35^(2) (ii) 1^(3) + 2^(3) + 3^(3) +…+ 20^(3) .

If a_(1),a_(2),a_(3),... are in A.P.and a_(i)>0 for each i,then sum_(i=1)^(n)(n)/(a_(i+1)^((2)/(3))+a_(i+1)^((1)/(3))a_(i)^((1)/(3))+a_(i)^((2)/(3))) is equal to

(-2-(1)/(3)i)^(3)

Evaluate: (i) {((-2)/(3))^(2)} (ii) [{((-1)/(3))^(2)}^(-2)]^(-1) (iii) {((3)/(2))^(-2)}