IIT JEE , AIEEE Forum - Mathematics , Algebra

sinjan.j

The value of (In i) + (In i)² + (In i)³ +.........+ (In i)ⁿ is

a) In i [ (n-1) In i + In ( i/e)]

b)n(n+1) In i

c) [( 2ⁿ -iⁿ Pi ⁿ) ( 2 pi i - pi² ) ] / 2ⁿ ( pi² + 4)

d) none of these

The answer is (c)

PLEASE POST ur DEATAIL SOLUTION

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iitkgp_bipin

Edited :

lni = i.(

/2)

given series is a GP with 1st term lni and common ratio lni.

so sum of GP = (lni).{(lni)ⁿ - 1} / (lni - 1)

If you put lni in this expression you'll find option (c) is correct.

For objective problems :

Put n=1 : the summation becomes lni.

Put n=1 in the given options of which (a) and (b) doesn't satisfy and (c) becomes lni for n=1. So correct option is (c)

divalli_oct07

nadeemoidu

Here 's my method,

(In i) + (In i)² + (In i)³ +.........+ (In i)ⁿ

ln(i) = i.(

/2)

So a= i.(

/2) and r = i.(

/2)

So the sum is i.(

/2) [ 1 - ( i

/2 )ⁿ ] / ( 1 - i

/2 )

Make the denominator real by multiplying it with its conjugate.

The answer is c.

And when there is an option "none of these" , how can u just substitute n=1 and say that it is "c"? It can be "d" also.

Mr.IITIAN007

Yeah , I go with Bipin's way.

ayshwarya

1st of all i dont agree wid dis quesn is it lnI i I if its ln i it wud b
2ln(-1) wich does not exist

rv_hbk

Its simple!!

i = cos(pi/2)+isin(pi/2)=e^i(pi/2)

Hence the given expression is:

i(pi/2) + (i(pi/2))²+ ....... upto nterms.

now, apply sum of g.p formula and get the ans.

ANS= C

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