For positive integers n1 and n2 , the value of the expression (1+i)n1+(i+13)n1+(i+15)n2+(i+17)n2 here i=√-1 is real if and only if
(A)n1=n2+1 (B) n1=n2-1
(C) n1=n2 (D)n1>0,n2 >0
pl post the soln as well .
591U can do it this way..
(1+i)n = C0 + C1i + C2i2 + C3i3 + ......
&(1-i)n= C0 - C1i + C2i2 - C3i3 + ......
when u add them all the terms with odd powers of i cancel off..and even powers of i are real..so
(1+i)n + (1-i)n is real for any n>0
Himanshu Giria thanks...481Is answer B?
I would prefer doing this by putting the options
(1+i)^n1 + (1-i)^n1 + (1+i)^n2 + (1-i)^n2
Taking B option as my answer and putting n2=2 and n1=1(n1,n2 are real)
we get
2 + (1+i)^2 + (1-i)^2= 2(which is real)
So,B
Anurag Ghosh Assuming this to be a single option answer,ididn't check d other options A and C which may be d answer....U can eliminate DHimanshu Giria How can u eliminate (D)
481Sry n1,n2 are positive integers.....make correction in my answer....
535(C)(1+i)n1+(1+i3)n1+(1+i5)n2+(1+i7)n2
= (1+i)n1+(1-i)n1+(1+i)n2+(1-i)n2
= (1+i)n1+2n1(1+i)n1+(1+i)n2+2n2(1+i)n2
=2n1.in1+2n1(1+i)n1+2n2.in2+2n2(1+i)n2
=2n1(in1+1)(1+i)n1+2n2(in2+1)(1+i)n2
Both the terms are symmetric wrt to each other. Therefore, n1=n2.
Also,
in the given expr.[(1+i)n1+(i+13)n1+(i+15)n2+(i+17)n2] just put n1=n2=1.
Himanshu Giria But the answer is given to be (D)Himanshu Giria moreover ,
if the terms are symmetric so how does n1=n2 make the last expression real...
Aditya Agarwal if the answer is d, then there is nothing to solve in the question, because d is itself given in the question!Aditya Agarwal if you replace n2 with n1, the solveed expression remains same, but the option given changes.
(A)n1=n2+1 becomes n2=n1+1 => n1=n2-1 which is option (B)
that is not possible.
that's why it should be C.
Aditya Agarwal ^ from this we can say that A and B are the same options.
now, if you substitute n1=2, n2=1
(1+i)n1+(1-i)n1+(1+i)n2+(1-i)n2
becomes (1+i)2 + (1-i)2 + (1+i) + (1-i)
= 2i -2i + 2 = 2 which is real.
if n1=1 n2=3. this eqn is still real. so the option is D
466In the 2nd , 3rd and 4th brackets it wud be 1+i instead of i+1 ,i.e., the powers 3,5,7 wud b on i ...
Himanshu Giria not on 1