IIT JEE , AIEEE Forum - Mathematics , Algebra

shinee

a)(1+i)²ⁿ /(1+i)²ⁿ⁺¹is a real no, then n=?

b)(

3 + i)ⁿ=(

3 -i)ⁿ , n belongs to N, then the least value of n is

c)sigma(0 to infinity)(2i/3)ⁿ=

Decoder

1) check it...no matter wat value of n ..it is always coming out to be complex..
it might be conjugate in the denominator..

for tht.. divide and multiply the numerator and denominator by root2^2n and root2^2n+1 respectively..
it gets u euler form ..solve to get the power of e as npi...
u will surely get the ans..

2)
divide both equations by 2^n ...

u get e ^ { i npi/6} = e ^ - {i npi/6}
thus it gets u
e^ix - e^-ix = 0 form..
which means sinx is zero..
thus npi/6 = mpi..
least value of n is 6...

Decoder

oh! i missed the third one..

just see the sequence..

1 + i { 2/3 + [2/3]^5 + [2/3]^9...} -i { [2/3]^2 + [2/3]^6...}

similarly those with i^4n+2..-1 wali form..& i^4n form..

all r infinite g.p solve them to get the ans..

shinee

canu solve the questions please
salute assured

shinee

salute assured 4 the one who solves it completely

ankitagg

is the answer to (c) is 1-3/2i

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